Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

\(=x^2-y^2\)

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

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

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

\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)

b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)

\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)

=0

Lời giải:
$H=(x^3-3x^2+3x-1)-(x^3+8)+3(x^2-16)$

$=x^3-3x^2+3x-1-x^3-8+3x^2-48$

$=(x^3-x^3)+(-3x^2+3x^2)+3x+(-1-8-48)$

$=3x-57=3.\frac{-1}{2}-57=\frac{-117}{2}$

chịu

a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)

\(=-4.\dfrac{1}{4}+10=9\)

b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)

\(=\left(-2\right).\left(32-32\right)=0\)

a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)

\(=4x^2-4x+1+9-4x^2\)

\(=-4x+10\)

\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

Khi  \(x=\frac{5}{4}\) \(\Leftrightarrow|2x+1|=2x+1\)

\(K=5\times\frac{5}{4}-\left(2\times\frac{5}{4}+1\right)-3\)

\(K=\frac{25}{4}-\frac{7}{2}-3\)

\(K=-\frac{1}{4}\)

Vậy  \(K=-\frac{1}{4}\Leftrightarrow x=\frac{5}{4}\)