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Bài này chắc phải giải theo kiểu lớp 7
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a}{3b}=\dfrac{3b}{4c}=\dfrac{4c}{5d}=\dfrac{5d}{2a}=\dfrac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)
\(\Rightarrow\left\{{}\begin{matrix}2a=3b\\3b=4c\\4c=5d\\5d=2a\end{matrix}\right.\)\(\Rightarrow2a=3b=4c=5d\)
\(\Rightarrow C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\)
\(=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}\)
\(=1+1+1+1\)
\(=4\)
Vậy \(C=4\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
Ta có:
Nếu:
\(\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\Leftrightarrow\left(2a+c\right)\left(b-d\right)=\left(a-c\right)\left(2b+d\right)\)
\(\Leftrightarrow2a\left(b-d\right)+c\left(b-d\right)=a\left(2b+d\right)-c\left(2b+d\right)\)
\(\Leftrightarrow2ab-2ad+bc-cd=2ab+ad-2bc+cd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\left(đpcm\right)\)
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3}{2}a\)
\(\dfrac{a}{2}=\dfrac{c}{5}\Rightarrow c=\dfrac{5}{2}a\)
=>B=\(\dfrac{a+7\cdot\left(\dfrac{3}{2}a\right)-2\cdot\left(\dfrac{5}{2}a\right)}{3a+2\cdot\left(\dfrac{3}{2}a\right)-\dfrac{5}{2}a}=\dfrac{a+\dfrac{21}{2}a-5a}{3a+3a-\dfrac{5}{2}a}=\dfrac{\dfrac{13}{2}a}{\dfrac{7}{2}a}=\dfrac{13}{7}\)
b)B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
B<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
B<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
B<\(1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)-\dfrac{1}{9}\)
B<1-\(\dfrac{1}{9}\)
B<\(\dfrac{8}{9}\)(1)
ta có:
B>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{10}\)
B>\(\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)...+\left(\dfrac{1}{9}+\dfrac{1}{9}\right)-\dfrac{1}{10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{10}\)
B>\(\dfrac{2}{5}\)
Câu 1:
a: \(\left(3x-15\right)=3^7:3^5\)
=>3x-15=9
=>3x=24
hay x=8
b: \(\left(4x+32\right)=43\cdot2^2\)
=>4x+32=172
=>4x=140
hay x=35
c: \(6^{2x-7}=216\)
=>2x-7=3
=>2x=10
hay x=5
d: \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x\cdot26=650\)
\(\Leftrightarrow5^x=25\)
hay x=2
\(\frac{2a}{a+b}+\frac{b}{a-b}=2< =>2\left(a-b\right)a+b\left(a+b\right)=2\left(a-b\right)\left(a+b\right).\)
\(< =>2a^2-2ab+ab+b^2=2a^2-2b^2\)
\(< =>3b^2-ab=0< =>b\left(3b-a\right)=0=>\orbr{\begin{cases}b=0\\3b-a=0\end{cases}}\)\(< =>\orbr{\begin{cases}b=0\\a=3b\end{cases}=>\orbr{\begin{cases}A=3\\A=1\end{cases}}}\)
Bài 1:
Ta có:
\(\left(100a+3b+1\right)\left(2^a+10a+b\right)=225\left(1\right)\)
Mà \(225\) lẻ nên \(\left\{{}\begin{matrix}100a+3b+1\\2^a+10a+b\end{matrix}\right.\) cùng lẻ \(\left(2\right)\)
\(*)\) Với \(a=0\) ta có:
Từ \(\left(1\right)\Leftrightarrow\left(100.0+3b+1\right)\left(2^a+10.0+b\right)=225\)
\(\Leftrightarrow\left(3b+1\right)\left(1+b\right)=225=3^2.5^2\)
Do \(3b+1\div3\) dư \(1\) và \(3b+1>1+b\)
Nên \(\left(3b+1\right)\left(1+b\right)=25.9\) \(\Rightarrow\left\{{}\begin{matrix}3b+1=25\\1+b=9\end{matrix}\right.\) \(\Leftrightarrow b=8\)
\(*)\) Với \(a\ne0\left(a\in N\right)\) ta có:
Khi đó \(100a\) chẵn, từ \(\left(2\right)\Rightarrow3b+1\) lẻ \(\Rightarrow b\) chẵn
\(\Rightarrow2^a+10a+b\) chẵn, trái với \(\left(2\right)\) nên \(b\in\varnothing\)
Vậy \(\left\{{}\begin{matrix}a=0\\b=8\end{matrix}\right.\)
Bài 2:
Ta có:
\(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+...+2017}\)
\(=\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}+\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}+...+\dfrac{1}{\dfrac{\left(1+2017\right).1009}{2}}\)
\(=\dfrac{2}{2.4}+\dfrac{2}{3.6}+\dfrac{2}{4.8}+...+\dfrac{2}{1009.2018}\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{1009.1009}\)
\(\Rightarrow A< \dfrac{1}{2.2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1008.1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\) (Đpcm)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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\(\dfrac{1}{2a-1}=\dfrac{2}{3b-1}=\dfrac{3}{4c-1}\Rightarrow\dfrac{2a-1}{1}=\dfrac{3b-1}{2}=\dfrac{4c-1}{3}\)
\(\Rightarrow\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}\)và 3a+2b-c=4
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}=\dfrac{36a-18+24b-8-12c+3}{18+16-9}=\dfrac{12\left(3a+2b-c\right)-23}{25}=\dfrac{12\cdot4-23}{25}=1\)
=>2a-1=1<=>a=1
3b-1=2<=>b=1
4c-1=3<=>c=1
Vậy...
thanks nhìu nha