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

\(=-7x\left(x^2-x-6\right)-10x^3+5x^2-20x+3x\left(x^2-5x-7\right)-\left(x^2-5x\right)\)

\(=-7x^3+7x^2+42x-10x^3+5x^2-20x+3x\left(x^2-5x-7\right)-x^2+5x\)

\(=-17x^3+11x^2+27x+3x^3-15x^2-21x\)

\(=-14x^3-4x^2+6x\)

`sqrt{2} = 1,414213...`

`sqrt{2} = 2/sqrt{2} `

\(\sqrt{2}=1,41421356...\)

\(M=-x^2+6x-11\\ =-\left(x^2-6x+9\right)+9-11\\ =-\left(x-3\right)^2-2\)

Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-3\right)^2\le0\forall x\\ \Rightarrow-\left(x-3\right)^2-2\le-2\forall x\\ \Rightarrow M\le-2\forall x\)

Dấu "=" xảy ra khi: \(x-3=0\Leftrightarrow x=3\)

Vậy \(M_{max}=-2\Leftrightarrow x=3\).

`M = -x^2 + 6x - 11`

`= -(x^2 - 6x + 11) `

`= -(x^2 - 2.3x + 3^2 + 2)`

`= -(x^2 - 2.3x + 3^2) - 2`

`= -(x-3)^2 - 2`

Do `(x-3)^2 ≥ 0`     `∀x` thuộc `R`

`=> -(x-3)^2 ≤ 0`    `∀x` thuộc `R`

`=> -(x-3)^2 - 2 ≤ -2`     ` ∀x` thuộc `R`

Hay `M ≤ -2` `∀x` thuộc `R`

Dấu `=` có khi: 

`(x-3)^2 = 0`

`<=> x - 3 = 0`

`<=> x = 3`

Vậy `M_(max) = -2 <=> x = 3`

\(\sqrt{\dfrac{2\left(4-\sqrt{7}\right)}{2}}-\sqrt{\dfrac{2\left(4+\sqrt{7}\right)}{2}}+\sqrt{2}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\sqrt{2}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}+\sqrt{2}=-\dfrac{2}{\sqrt{2}}+\dfrac{2}{\sqrt{2}}=0\)

\(\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{\dfrac{8+2\sqrt{7}}{2}}+\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)

\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}\)

\(=\dfrac{-2}{\sqrt{2}}+\sqrt{2}=0\)

a: 

ĐKXĐ: \(x\notin\left\{1;-3\right\}\)

\(A=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)

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

\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\)

\(=\dfrac{x^2-x}{\left(x-1\right)}\cdot\dfrac{1}{x+3}=\dfrac{x}{x+3}\)

b: |x-5|=2

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Khi x=7 thì \(A=\dfrac{7}{7+3}=\dfrac{7}{10}\)

c: Để A nguyên thì \(x⋮x+3\)

=>\(x+3-3⋮x+3\)

=>\(-3⋮x+3\)

=>\(x+3\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{-2;-4;0;-6\right\}\)

\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\ < =>\left[\left(x+2\right)-\left(2x-3\right)\right]^2=25\\ < =>\left(x+2-2x+3\right)^2-25=0\\ < =>\left(-x+5\right)^2-5^2=0\\ < =>\left(-x+5-5\right)\left(-x+5+5\right)=0\\ < =>-x\left(-x+10\right)=0\\ < =>x\left(x-10\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

Vậy: ... 

\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\ \Leftrightarrow\left(x+2-2x+3\right)^2=5^2\\\Leftrightarrow\left(-x+5\right)^2=5^2\\ \Leftrightarrow\left[{}\begin{matrix}-x+5=5\\-x+5=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy...

Ta có:

\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\\ =\left(y^2-5y+8y-40\right)-\left(y^2+4y-y-4\right)\\ =y^2+3y-40-y^2-3y+4\\ =-36\)

=> Giá trị của bt không phụ thuộc vào biến y

\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)

\(=y^2+8y-5y-40-\left(y^2-y+4y-4\right)\)

\(=y^2+3y-40-y^2-3y+4\)

=-36

=>M không phụ thuộc vào biến

`2x^3-3x^2-32x-15`

`=2x^3-10x^2+7x^2-35x+3x-15`

`=2x^2(x-5)+7x(x-5)+3(x-5)`

`=(x-5)(2x^2+7x+3)`

`=(x-5)(2x^2+x+6x+3)`

`=(x-5)[x(2x+1)+3(2x+1)]`

`=(x-5)(2x+1)(x+3)`

\(P=2x^4+3x^2y^2+y^4+y^2\)

\(=2x^4+2x^2y^2+x^2y^2+y^4+y^2\)

\(=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)

\(=2x^2+y^2+y^2=2\left(x^2+y^2\right)=2\)

Ta có :

\(P\left(x\right)=2x^4+3x^2y^2+y^4+y^2\)

\(\Rightarrow P\left(x\right)=x^4+2x^2y^2+y^4+x^4+x^2y^2+y^2\)

\(\Rightarrow P\left(x\right)=\left(x^2+y^2\right)^2+x^2\left(x^2+y^2\right)+y^2\)

\(\Rightarrow P\left(x\right)=1^2+x^2.1+y^2\) Vì \(\left(x^2+y^2=1\right)\)

\(\Rightarrow P\left(x\right)=1^2+x^2+y^2=1+1=2\)

Vậy \(P\left(x\right)=2\)

\(0< a< 2\Rightarrow a\left(a-2\right)< 0\Rightarrow a^2< 2a\)

Tương tự: \(\left\{{}\begin{matrix}b\left(b-2\right)< 0\\c\left(c-2\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b^2< 2b\\c^2< 2c\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2< 2\left(a+b+c\right)\)

\(\Rightarrow a^2+b^2+c^2< 2.3=6\)