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chẳng hài j cả

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0 vui cũng 0 bùn

=v

a)

\(A=\left(x^2-4x+4\right)-11\)

\(=\left(x-2\right)^2-11\)

Ta có

\(\left(x-2\right)^2-11\ge-11\)

Dấu " = " xảy ra khi x = 2

Vậy MIN_A= - 11 khi x=2

b) 

\(B=-\left(x^2-4x+4\right)-3\)

\(B=-\left(x-2\right)^2-3\)

Ta có

\(-\left(x-2\right)^2-3\le-3\) với mọi x

Dấu " = " xảy ra khi = 2

Vậy MAX_B= - 3 khi x = 2

\(M=5^3\left(1+5+5^2\right)\)

\(M=5^3\left(5^3-1\right)\)

\(M=5^6-5^3\)

\(M=15500\)

văn nghị luận ak

\(\left(x-1\right)\left(x+2\right)-x^2=5\)

\(\Leftrightarrow x^2+2x-x-2-x^2=5\)

\(\Leftrightarrow x-2=5\)

\(\Leftrightarrow x=7\)

Ta có

\(\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{100}+\left(2-5\right)^{688}\)

\(=\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{100}+\left(-3\right)^{688}\)

\(=\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{100}+3^{688}\)

\(\Rightarrow\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{100}+3^{688}=0\)

\(\Rightarrow\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{100}=-3^{688}\)

Mà \(\begin{cases}\left(x-\frac{1}{5}\right)^{2014}\ge0\\\left(y+0,4\right)\ge0\end{cases}\)  với mọi x; y ( vì có số mũ chẵn )

\(\Rightarrow x;y\in\varnothing\)

Chia cả tử và mẫu của mỗi phân số tương ứng cho b²⁰¹⁵; b²⁰¹⁴

=> cần chứng minh: \(\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}>\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}\)

Ta có: \(VT=\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}=\frac{\left(\frac{a}{b}\right)^{2015}+1}{\left(\frac{a}{b}\right)^{2015}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}\)

\(VP=\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}=\frac{\left(\frac{a}{b}\right)^{2014}+1}{\left(\frac{a}{b}\right)^{2014}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)

Vì a> b > 0 => a/b  > 1. Do đó:

\(\left(\frac{a}{b}\right)^{2015}+1>\left(\frac{a}{b}\right)^{2014}+1\)

=> \(\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)

=> VT > VP