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\(A=\left(1+2+3+...+100\right)^2\cdot\left(a+2b\right)^4\cdot\left(a+3b\right)^5\cdot\left(\dfrac{4}{5}-\dfrac{4}{5}\right)^6\)
=0
a) Ta có: \(a=\frac{3}{5}=0,6\)
\(A=\left(1^2+2^2+3^2+...+20^2\right).\left(a+b\right)\left(2a+b\right)\left(a+3b\right)\)
\(\Rightarrow A=\left(1^2+2^2+3^2+...+20^2\right)\left(a+b\right)\left(2a+b\right)\left[0,6+3.\left(-0,2\right)\right]\)
\(\Rightarrow A=\left(1^2+2^2+...+20^2\right)\left(a+b\right)\left(2a+b\right)\left(0,6-0,6\right)\)
\(\Rightarrow A=\left(1^2+2^2+...+20^2\right)\left(a+b\right)\left(2a+b\right).0\)
\(\Rightarrow A=0\)
Vậy A = 0
b) Ta có: \(\frac{a}{b}=\frac{3}{4}\Rightarrow\frac{a}{3}=\frac{b}{4}\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k\Rightarrow\left\{\begin{matrix}a=3k\\b=4k\end{matrix}\right.\)
\(B=\frac{2a-3b}{a-3b}=\frac{2.3.k-3.4.k}{3k-3.4.k}=\frac{6k-12k}{3k-12k}=\frac{\left(6-12\right)k}{\left(3-12\right)k}=\frac{-6}{-9}=\frac{2}{3}\)
Vậy \(B=\frac{2}{3}\)
\(2a^2+2b^2=5ab\)
<=> \(2a^2+2b^2-5ab=0\)
<=> \(2a^2-4ab-ab+2b^2=0\)
<=> \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)
<=> \(\orbr{\begin{cases}2a=b\\a=2b\end{cases}}\)
Do b > a > 0
=> b = 2a
\(A=\frac{a+b}{a-b}=\frac{a+2a}{a-2a}=\frac{3a}{-a}=-3\)
\(2a^2+2b^2=5ab\)
<=> \(2a^2+2b^2-5ab=0\)
<=> \(2a^2-4ab-ab+2b^2=0\)
<=> \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)
<=> \(\left(2a-b\right)\left(a-2b\right)=0\)
<=> \(\orbr{\begin{cases}2a-b=0\left(L\right)\\a-2b=0\end{cases}}\)
=> \(a=2b\)
=> \(A=\frac{a+2b}{2a-b}=\frac{2b+2b}{2.2b-b}=\frac{4b}{3b}=\frac{4}{3}\)