\(-3x^2-7x+10\)

\(=3x-3x^2+10-10x\)

\(=3x.\left(1-x\right)+10.\left(1-x\right)=\left(3x+10\right).\left(1-x\right)\)

\(-3x^2+3y^2-4xz-4yz\)

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

\(=3\left(y-x\right)\left(x+y\right)-4z\left(x+y\right)\)

\(=\left(x+y\right)\left(3y-3x-4z\right)\)

\(\left(x^2+x-2\right)^2=3\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left[\left(x-1\right)\left(x+2\right)\right]^2=3\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)^2=3\left(x^4+x^2+1\right)\)

\(\Leftrightarrow x^4+4x^3+4x^2-2x^3-8x^2-8x+x^2+4x+4=3x^4+3x^2+3\)

\(\Leftrightarrow x^4+2x^3-3x^2-4x+4-3x^4-3x^2-3=0\)

\(\Leftrightarrow-2x^4+2x^3-6x^2-4x+1=0\)

\(x+x^4\)

\(=x\left(1+x^3\right)\)

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

x(1+x³

^{dấu ngoặc đằng sau ko viết được}

xem trên mạng nhé 

trả lòi làm dell j

Ta có:

\(\sqrt{x^2-1}\)

\(=\sqrt{\frac{1}{4}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)^2-1}\)

\(=\sqrt{\frac{1}{4}\left(\frac{a}{b}+2+\frac{b}{a}\right)-1}\)

\(=\sqrt{\frac{\left(a-b\right)^2}{4ab}}\)

\(=\frac{|a-b|}{2\sqrt{ab}}\)

Thế vào Q ta được:

\(Q=\frac{\frac{2ab|a-b|}{2\sqrt{ab}}}{\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)-\frac{|a-b|}{2\sqrt{ab}}}\)

\(=\frac{2ab|a-b|}{\left(a+b\right)-|a-b|}\)

Vì \(|a-b|=\hept{\begin{cases}a-b\left(a\ge b\right)\\b-a\left(a< b\right)\end{cases}}\)

\(\Rightarrow Q=\hept{\begin{cases}a-b\left(a\ge b\right)\\\frac{b}{a}\left(b-a\right)\left(a< b\right)\end{cases}}\)

xem trên mạng nhé 

\(Q=\frac{x^2+2x+1-x-1+1}{x^2+2x+1}=1+\frac{1}{x+1}+\frac{1}{\left(x+1\right)^2}\)

Đặt \(A=\frac{1}{x+1}.\text{ta có: }Q=A+A^2+1=A^2+\frac{2A.1}{2}+\frac{1}{4}+\frac{3}{4}=\left(A+\frac{1}{2}\right)^2\ge\frac{3}{4}\)

\(\text{dấu bằng xảy ra khi: }A=\frac{1}{2}.\Rightarrow\frac{1}{x+1}=\frac{1}{2}\Rightarrow x=1.\text{Vậy}...\)