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Từ \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}=2\left(a^{101}+b^{101}\right)\)
\(\Rightarrow a^{100}+b^{100}+a^{102}+b^{102}-2\left(a^{101}+b^{101}\right)=0\)
\(\Rightarrow\left(a^{102}-2a^{101}+a^{100}\right)+\left(b^{102}-2b^{101}+b^{100}\right)=0\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2=0\left(1\right)\)
Vif \(\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2\ge0\forall a\\\left(b^{51}-b^{50}\right)^2\ge0\forall b\end{cases}}\)
\(\Rightarrow\left(a^{51}-a^{50}\right)^2+\left(b^{51}-b^{50}\right)^2\ge0\forall a,b\left(2\right)\)
Tứ (1) và (2) :
\(\Rightarrow\hept{\begin{cases}\left(a^{51}-a^{50}\right)^2=0\\\left(b^{51}-b^{50}\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}-a^{50}=0\\b^{51}-b^{50}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^{51}=a^{50}\\b^{51}=b^{50}\end{cases}}\)
Vì a,b là các số thực dương nên \(a=b=1\)
\(\Rightarrow P=a^{2007}+b^{2007}=1^{2007}+1^{2007}=1+1=2\)
Vậy \(P=2\)
đây là bài tổng quát nè bạn, áp dụng bài này nhé ^_^
https://olm.vn/hoi-dap/question/1123004.html
Các bợn làm nhanh dùm mk nha. Bài kiểm tra sáng mai mình nộp rồi. Ai nhanh nhất mình tick cho nha
sai đề nha bn : là \(\dfrac{2}{\left(a+b\right)^{1002}}\) mới đúng
+ \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}=\dfrac{x^2+y^2}{a+b}\)
\(\Rightarrow\dfrac{bx^4+ay^4}{ab}=\dfrac{x^2+y^2}{a+b}\)
\(\Rightarrow\left(bx^4+ay^4\right)\left(a+b\right)=ab\left(x^2+y^2\right)\)\(\Rightarrow abx^4+aby^4+a^2y^4+b^2x^4=abx^2+aby^2\)
\(\Rightarrow a^2y^4+b^2x^4=abx^2\left(1-x^2\right)+aby^2\left(1-y^2\right)\)
\(\Rightarrow a^2y^4+b^2x^4=abx^2y^2+abx^2y^2\)
\(\Rightarrow\left(ay^2\right)^2+\left(bx^2\right)^2-2abx^2y^2=0\)
\(\Rightarrow\left(ay^2-bx^2\right)^2=0\)
\(\Rightarrow ay^2-bx^2=0\Rightarrow ay^2=bx^2\)
\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) ( tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}=\dfrac{y^{2002}}{b^{1002}}=\dfrac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}+\dfrac{y^{2004}}{b^{1002}}=\dfrac{2}{\left(a+b\right)^{1002}}\) ( đpcm )
a. \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Rightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Rightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)
\(\Rightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Rightarrow x-105=0\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
\(\Rightarrow x=105\)
b. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)
\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
\(\Rightarrow x=50\)
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
Dễ dàng thấy nhân tử thứ hai luôn bé thua 0 nên \(x-105=0\)\(\Leftrightarrow x=105\)
b) Kĩ thuật làm tương tự câu a cộng mỗi phân số VT với 1 thì VP=0 và ta có nhân tử chung 50-x
Vì a100+ b100; a101 + b101 ;a102 + b102 đều = nhau nên a chỉ có thể = 1 => a2010 +b2010 = 12010+12010 = 1+1 = 2
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
với a, b >0
\(a^9+b^9=a^{10}+b^{10}< =>a^9\left(a-1\right)+b^9\left(b-1\right)=0\)
\(a^{10}+b^{10}=a^{11}+b^{11}< =>a^{10}\left(a-1\right)+b^{10}\left(b-1\right)=0\)
trừ vế theo vế ta được (a-1)(a10-a9) + (b-1)(b10-b9) = 0 <=> [b3(b-1)]2 + [b3(b-1)]2 =0
<=> \(\hept{\begin{cases}a^3\left(a-1\right)=0\\b^3\left(b-1\right)=0\end{cases}< =>\hept{\begin{cases}a-1=0\\b-1=0\end{cases}< =>}}\)a = b =1
vậy P= 2020
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
Mà đẳng thức trên xảy ra dấu =
\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)
Bài kia tí nghĩ nốt, khó v
Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)
