giải pt: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:

+ ĐK:\(\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8-6\sqrt{x-1}\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

+ pt đã cho \(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\) (*)

Th1: \(\left\{{}\begin{matrix}\sqrt{x-1}-2< 0\\\sqrt{x-1}-3< 0\end{matrix}\right.\)

(*) \(\Leftrightarrow2-\sqrt{x-1}+3-\sqrt{x-1}=1\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(N\right)\)

Th2: \(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\ge0\end{matrix}\right.\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-1}-3=1\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(N\right)\)

Th3: \(\sqrt{x-1}-3< 0\le\sqrt{x-1}-2\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+3-\sqrt{x-1}=1\Leftrightarrow1=1\left(đúng\right)\)

Kl: \(x\ge1\)

+Tuấn 10B_2 (T ko biết đánh word nên dùng tạm .V)

GPT: \(\(\sqrt{x+3}+\sqrt[3]{x}=3\)\) (Bài này cách lp 9 dễ t ko giải nữa)

Vì \(\(f\left(x\right)=\sqrt{x+3}+\sqrt[3]{x}=3\)\) là hàm tăng trên tập [-3;\(\(+\infty\)\))

Ta có: Nếu \(\(x&gt;1\Leftrightarrow f\left(x\right)&gt;f\left(1\right)=3\)\)nên pt vô nghiệm

Nếu \(\(-3\le x&lt; 1\Leftrightarrow f\left(x\right)&lt; f\left(1\right)=3\)\)nên pt vô nghuêmj

Vậy x = 1

B2, GHPT: \(\(\hept{\begin{cases}2x^2+3=\left(4x^2-2yx^2\right)\sqrt{3-2y}+\frac{4x^2+1}{x}\\\sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\end{cases}}\)\)

ĐK \(\(\hept{\begin{cases}-\frac{1}{2}\le y\le\frac{3}{2}\\x\ne0\\x\ne-\frac{1}{2}\end{cases}}\)\)

Xét pt (1) \(\(\Leftrightarrow2x^2+3-4x-\frac{1}{x}=x^2\left(4-2y\right)\sqrt{3-2y}\)\)

\(\(\Leftrightarrow-\frac{1}{x^3}+\frac{3}{x^2}-\frac{4}{x}+2=\left(4-2y\right)\sqrt{3-2y}\)\)

\(\(\Leftrightarrow\left(-\frac{1}{x}+1\right)^3+\left(-\frac{1}{x}+1\right)=\left(\sqrt{3-2y}\right)^3+\sqrt{3-2y}\)\)

Xét hàm số \(\(f\left(t\right)=t^3+t\)\)trên R có \(\(f'\left(t\right)=3t^2+1&gt;0\forall t\in R\)\)

Suy ra f(t) đồng biến trên R . Nên \(\(f\left(-\frac{1}{x}+1\right)=f\left(\sqrt{3-2y}\right)\Leftrightarrow-\frac{1}{x}+1=\sqrt{3-2y}\)\)

Thay vào (2) \(\(\sqrt{2-\left(1-\frac{1}{x}\right)}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\sqrt{\frac{1}{x}+1}=\frac{\sqrt[3]{x^2\left(x+2\right)}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\left(2x+1\right)\sqrt{\frac{1}{x}+1}=x+2+\sqrt[3]{x^2\left(x+2\right)}\)\)

\(\(\Leftrightarrow\left(2+\frac{1}{x}\right)\sqrt{1+\frac{1}{x}}=1+\frac{2}{x}+\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow f\left(\sqrt{1+\frac{1}{x}}\right)=f\left(\sqrt[3]{1+\frac{2}{x}}\right)\)\)

\(\(\Leftrightarrow\sqrt{1+\frac{1}{x}}=\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow\left(1+\frac{1}{x}\right)^3=\left(1+\frac{2}{x}\right)^2\)\)

Đặt \(\(\frac{1}{x}=a\)\)

\(\(\Rightarrow Pt:\left(a+1\right)^3=\left(2a+1\right)^2\)\)

Tự làm nốt , mai ra lớp t giảng lại cho ...

    \(\sqrt{x^2+8}-7x=\sqrt{x^2+3}-6\)(1)

\(\Leftrightarrow\sqrt{x^2+8}-3=7x-7+\sqrt{x^2+3}-2\)

\(\Leftrightarrow\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{\left(\sqrt{x^2+3}-2\right)\left(\sqrt{x^2+3}+2\right)}{\sqrt{x^2+3}+2}\)

\(\Leftrightarrow\frac{x^2+8-9}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+3}+2}\)

\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+8}+3}-7\left(x-1\right)-\frac{x^2-1}{\sqrt{x^2+3+2}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}\right)=0\)

\(\Leftrightarrow x-1=0\)

hay \(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}=0\)(2)

Từ (1), có:

\(\sqrt{x^2+8}-\sqrt{x^2+3}=7x-6>0\)

\(\Leftrightarrow7x-6>0\)

\(\Leftrightarrow x>\frac{6}{7}\)

Khi đó, có:

     \(\frac{x+1}{\sqrt{x^2+8}+3}-\frac{\sqrt{x+1}}{\sqrt{x^2+3}+2}<0\)

\(\Rightarrow\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+3}+2}-7<0\)

Vậy, pt (2) vô nghiệm

Do đó, pt (1) có 1 nghiệm là x = 1

\(\sqrt{5-3x}+\sqrt{x+1}=\sqrt{3x^2-4x+4}\)

\(\Leftrightarrow5-3x+x+1+2\sqrt{\left(5-3x\right)\left(x+1\right)}=3x^2-4x+4\)

\(\Leftrightarrow2\sqrt{-3x^2+2x+5}=3x^2-2x-2\)

\(\Leftrightarrow2\sqrt{-\left(3x^2-2x-5\right)}=3x^2-2x-5+3\)

Đặt \(3x^2-2x-5=t\left(t\le0\right)\)

\(\Rightarrow t=-1\)

\(\Leftrightarrow3x^2-2x-5=-1\)

\(\Leftrightarrow3x^2-2x-4=0\)

\(\Leftrightarrow x=\frac{1+\sqrt{13}}{3}\)

Kiểm duyệt giùm

Giải các pt sau:

a)\(\left|3x+1\right|=\left|x+1\right|\)

b)\(\left|x^2-3\right|=\left|x-\sqrt{3}\right|\)

c)\(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)

d)\(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)

e) \(\left|x^2-1\right|+\left|x+1\right|=0\)

f)\(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)

g) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)

h) \(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)

Mọi người giúp em gấp với!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Giải các pt sau:

a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)

b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)

c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)

d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)

f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)

g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)

\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)

\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)

\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)

Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)

\(\Rightarrow\)PT vô nghiệm

a,

ta có:

(x²+7x+3)²=x⁴+14x³+55x²+42x+9

(8x+4)(x²+5x+2)=8x³+44x²+36x+8

=>x⁴+14x³+55x²+42x+9=8x³+44x²+36x+8

<=>x⁴+6x³+11x²+6x+1=0

xét x=0 ko phải no của pt

xét x khác 0

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+6\left(x+\frac{1}{x}\right)+11=0\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+6\left(x+\frac{1}{x}\right)+9=0\Leftrightarrow\left(x+\frac{1}{x}+3\right)^2=0\Rightarrow x=\frac{-3+\sqrt{5}}{2};\frac{-3-\sqrt{5}}{2}\)

d,

xét n=1=> mệnh đề luôn đúng

giả sử mệnh đề đúng với n=k

ta sẽ cm nó đúng với n=k+1

với n=k+1

=>(n+1)(n+2)..(n+n)=2n(n+1)(n+2)...(2n-1)

=2(k+1)(k+2).....2k chia hết cho 2^k+1

=>(n+1)(n+2)(n+3)...(n+n) chia hết cho 2ⁿ

c,

ta có:

\(\left(1+x\right)\left(1+\frac{y}{x}\right)=1+x+y+\frac{y}{x}\ge1+y+2\sqrt{y}=\left(\sqrt{y}+1\right)^2\)

\(\Rightarrow\left(1+x\right)\left(1+\frac{y}{x}\right)\left(1+\frac{9}{\sqrt{y}}\right)^2\ge\left[\left(\sqrt{y}+1\right)\left(1+\frac{9}{\sqrt{y}}\right)\right]^2\)

\(=\left(\sqrt{y}+\frac{9}{\sqrt{y}}+10\right)^2\ge\left(6+10\right)^2=256\left(Q.E.D\right)\)

dấu = xảy ra khi y=9;x=3

b,

x⁷+xy⁶=y¹⁴+y⁸

<=>(x⁷-y¹⁴)+(xy⁶-y⁸)=0

<=>(x-y²)(x+y²)+y⁶(x-y²)=0

<=>(x-y²)(x+y²+y⁶)=0

xét x=y²

\(\Rightarrow\sqrt{4x+5}+\sqrt{y^2+8}=\sqrt{4y^2+5}+\sqrt{y^2-1}\)

\(\Rightarrow\sqrt{4y^2+5}+\sqrt{y^2+8}=6\)

\(\Rightarrow\left(\sqrt{4y^2+5}-3\right)+\left(\sqrt{y^2+8}-3\right)=0\)

\(\Rightarrow\frac{4y^2-4}{\sqrt{4y^2+5}+3}+\frac{y^2-1}{\sqrt{y^2+8}+3}=0\)

\(\Rightarrow\left(y^2-1\right)\left(\frac{4}{\sqrt{4y^2+5}+3}+\frac{1}{\sqrt{y^2+8}+3}\right)=0\)

\(\frac{4}{\sqrt{4y^2+5}+3}+\frac{1}{\sqrt{y^2+8}+3}>0\Rightarrow y^2=1\Rightarrow\left(x;y\right)=\left(1;1\right);\left(1;-1\right)\)

xét x+y²+y⁶=0

<=>x=-y²-y⁶

lại có:

x⁷+xy⁶=y¹⁴+y⁸

<=>x(x⁶+y⁶)=y¹⁴+y⁸

<=>-(y²+y⁶)(x⁶+y⁶)=y¹⁴+y⁸

mà \(-\left(y^2+y^6\right)\left(x^6+y^6\right)\le0\le y^{14}+y^8\)

<=>y=0=>x=0(ko thỏa mãn)

vậy nghiệm của pt:(x;y)=(1;-1);(1;1)