\(\left\{{}\begin{matrix}-\left|5x-2\right|\le0\\-\left|3g+12\right|\le0\end{matrix}\right.\Rightarrow-\left|5x-2\right|-\left|3g+12\right|\le0\)

\(\Rightarrow A=4-\left|5x-2\right|-\left|3g+12\right|\le4\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left|5x-2\right|=0\\\left|3g+12\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\g=-4\end{matrix}\right.\)

Vậy \(MAX_A=4\) khi \(\left\{{}\begin{matrix}x=\dfrac{2}{5}\\g=-4\end{matrix}\right.\)

\(-\left|10,2-3,x\right|\le0\)

\(\Rightarrow A=-\left|10,2-3,x\right|-14\le-14\)

Dấu " = " khi \(-\left|10,2-3,x\right|=0\Rightarrow10,2=3,x\)

\(\Rightarrow102=3x\Rightarrow x=34\)

Vậy \(MAX_A=-14\) khi x = 34

Bài 1:

a, Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(A=\left|x-3\right|+\left|x-5\right|=\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|=\left|2\right|=2\)

Dấu " = " khi \(\left\{{}\begin{matrix}x-3\ge0\\5-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\)

Vậy \(MIN_A=2\) khi \(3\le x\le5\)

b, Ta có: \(\left\{{}\begin{matrix}\left|y^2-25\right|\ge0\\\left|x^2-4\right|\ge0\end{matrix}\right.\Rightarrow\left|y^2-25\right|+\left|x^2-4\right|\ge0\)

\(\Rightarrow B\ge3\)

Dấu " = " khi \(\left\{{}\begin{matrix}y^2-25=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\pm5\\x=\pm2\end{matrix}\right.\)

Vậy \(MIN_B=2\) khi \(\left\{{}\begin{matrix}x=\pm2\\y=\pm5\end{matrix}\right.\)

Bài 2:

a, Xét \(x\ge-2\) có:

\(A=3x-3x-6-12=-18\)

+) Xét x < -2 có:

\(A=3x+3x+6-12=6x-6\)

Vậy...

b, tương tự

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Mà \(\left\{{}\begin{matrix}4\left(x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}4\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\)

Ta có: \(M=\left(x+y\right)^{2017}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\)

\(=\left(-1\right)^{2008}=1\)

Vậy M = 1

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

\(=\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

\(=\dfrac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}\)

\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}\)

\(=\dfrac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}\)

\(=\dfrac{8}{1-x^8}\)

\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)

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

\(=\left(3x^2-1\right)\left(2x^2-3\right)\)

\(G=\left(5x-7\right)\left(7x+3\right)-\left(7x+2\right)\left(5x-4\right)\)

\(=35x^2+15x-49x-21-35x^2+28x-10x+8\)

\(=-16x-13\)

Thay x = -3 \(\Rightarrow G=35\)

Vậy G = 35 khi x = -3

\(F=4.x.2-2.x+3x\left(x-5\right)\)

\(=6x+3x^2-15x\)

\(=3x^2-9x=3x\left(x-9\right)\)

Thay x = -1 có:

\(F=-3.\left(-10\right)=30\)

Vậy...

a, \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21⋮7\)

\(\Rightarrowđpcm\)

b, \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

Bài 3:

Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\\\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\\\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\end{matrix}\right.\Rightarrow P\ge\dfrac{a+b+c}{a+b+c}=1\) (1)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\\\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\\\dfrac{c}{c+a}< \dfrac{b+c}{a+b+c}\end{matrix}\right.\Rightarrow P< \dfrac{2a+2b+2c}{a+b+c}=2\)(2)

Từ (1), (2) \(\Rightarrow1< P< 2\)

\(\Rightarrow P\notin N\)

\(\Rightarrowđpcm\)