đk: \(x\ge-1\)

-xét x bằng 0 (tm)

-xét x khác 0=>phương trình có nghiệm khi x<0,khi đó ta có:

\(x+2.\sqrt{2.x^2.\left(x+1\right)}=0\) mà x < 0 nên khi rút gọn cho x ta có:

\(1-2.\sqrt{2\left(x+1\right)}=0\) => giải ra ta có  x=\(\frac{-7}{8}\) (tm).     vậy phương trình có 2 nghiệm là 0 và\(\frac{-7}{8}\)

B=5x²+4xy-2(x-2y)+2y²+3

=5x²+4xy-2x+4y+2y²+3

=(4x²+4xy+y²)+(x²-2x+1)+(y²+4y+4)-2

=(2x+y)²+(x-1)²+(y+2)²-2  \(\ge\) -2

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+y=0\\x-1=0\\y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)

thanks b

Ta có : \(4x^2-5x+2\)

\(=\left(2x\right)^2-2.2x.\frac{5}{4}+\frac{25}{16}+\frac{7}{16}\)

\(=\left(2x-\frac{5}{4}\right)^2+\frac{7}{16}\)

Do \(\left(2x-\frac{5}{4}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{5}{4}\right)^2+\frac{7}{16}\ge\frac{7}{16}>0\forall x\left(đpcm\right)\)

Học tốt nha bạn 

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

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

\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

Đặt \(a=x^2-5x+5\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)

\(\Leftrightarrow A=a^2-1^2+1\)

\(\Leftrightarrow A=a^2\)

Thay \(a=x^2-5x+5\)vào A ta có :

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

b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)

\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)

\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)

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

Làm tương tự câu a)

c) \(12x^2-3xy-8xz+2yz\)

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

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

\(B=\left(x^2-x\right)\left(x^2-x-6\right)+9\)

\(=\left(x^2-x\right)^2-6\left(x^2-x\right)+9=\left(x^2-x-3\right)^2\ge0\)

Vậy GTLN của B là 0

Xin lỗi bạn, GTNN của A là 0 nhé.

\(\left(m^2+n^2-5\right)^2-4\left(mn+2\right)^2\)

\(=\left(m^2+n^2-5\right)^2-\left(2mn+4\right)^2\)

\(=\left(m^2+n^2-5-2mn-4\right)\left(m^2+n^2-5+2mn+4\right)\)

\(=\left[\left(m^2-2mn+n^2\right)-3^2\right].\left[\left(m^2+2mn+n^2\right)-1^2\right]\)

\(=\left[\left(m-n\right)^2-3^2\right]\left[\left(m+n\right)^2-1^2\right]\)

\(=\left(m-n-3\right)\left(m-n+3\right)\left(m+n-1\right)\left(m+n+1\right)\)

\(\left(a-2\right)\left(a+2\right)\left(a^2+4\right)-\left(a^2+5\right)\left(a^2-5\right)\)

\(=\left(a^2-4\right)\left(a^2+4\right)-\left(a^2+5\right)\left(a^2-5\right)\)

\(=a^4-16-a^4+25\)

\(=9\)