• \(6x^2+12x+6=\left(6x^2+6x\right)+\left(6x+6\right)\)

\(=6x\left(x+1\right)+6\left(x+1\right)\)

\(=6\left(x+1\right)\left(x+1\right)\)

\(=6\left(x+1\right)^2\)

• \(8x^2-23x-3=\left(8x^2-24x\right)+\left(x-3\right)\)

\(=8x\left(x-3\right)+\left(x-3\right)\)

\(=\left(8x+1\right)\left(x-3\right)\)

*)(x+ 2)² = (2x - 1)²

<=>(x+2)²-(2x-1)²=0

<=>(x+2-2x+1)(x+2+2x-1)=0

<=>(3-x)(3x+1)=0

<=>x=3 hoặc x=-1/3

*) (x+2)² - x² + 4 = 0

<=>x²+4x+4-x²+4=0

<=>4x+8=0

<=>4.(x+2)=0

<=>x=-2

*) (x² - 2)² + 4(x - 1)² - 4(x² - 2)(x - 1) = 0

<=>[(x²-2)-2.(x-1)]²=0

<=>(x²-2-2x+1)²=0

<=>x²-2x-1=0 (nghiệm lẻ quá bạn xem lại nhé)

\(A=\frac{x+y}{z}+1+\frac{x+z}{y}+1+\frac{y+z}{x}+1-3\)

\(A=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3\)

\(A=\left(x+y+z\right)\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=\left(z+y+z\right)\cdot0-3=-3\)

Vậy, A = -3

cảm ơn bạn nha

\(\sqrt{x^2+2}\ge\sqrt{2}>1\forall x.\)

\(\Rightarrow\sqrt{x^2+2}+\sqrt{x^2+2x+3}>1\forall x\)

Dấu "=" không xảy ra nên PT vô nghiệm.

chữ bị lỗi .... ~0~

1/

a/  \(x^2+y^2=x^2+y^2+2xy-2xy\)\(=\left(x+y\right)^2-2xy\)

thay vào: \(\left(x+y\right)^2-2xy=a^2-2b\)

b/ \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+y^2+2xy-xy-2xy\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

thay vào:  \(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=a\left(a^2-3b\right)\)

c/ \(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2\)

thay vào: \(\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)

a) \(-x^2+6x+1=-\left(x^2-6x+9\right)+10=-\left(x-3\right)^2+10\le10\)

Vậy Max = 10 <=> x = 3

b) \(-5x^2-4x+1=-5\left(x^2+2.x.\frac{2}{5}+\frac{4}{25}\right)+\frac{4}{5}+1=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)

Vậy Max = \(\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)