a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

A)    \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)

           \(x=\dfrac{22}{15}\)

b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)

C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)

\(a,\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{2}\\2x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-\dfrac{1}{2}\\2x=-\dfrac{1}{2}-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\2x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)  \(b,\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\\ \Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\\ \Leftrightarrow\left(\dfrac{3+4}{6}\right)x=\dfrac{-10}{3}+\dfrac{3}{3}\\ \Leftrightarrow\dfrac{7}{6}x=\dfrac{-7}{3}\\ \Leftrightarrow x=\left(-\dfrac{7}{3}\right):\dfrac{7}{6}\\ \Leftrightarrow x=-2\)

Vậy \(x=0;x=-\dfrac{1}{2}\)                 Vậy \(x=-2\)

\(c,\dfrac{x-12}{4}=\dfrac{1}{2}\\ \Leftrightarrow2.\left(x-12\right)=4\\ \Leftrightarrow2x-24=4\\ \Leftrightarrow2x=24+2\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=26:2=13\)

Vậy \(x=13\)

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6\)

\(\Leftrightarrow-4x+13=6\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

\(b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=10\)

\(\Leftrightarrow x^2+6x+9+16-x^2=10\)

\(\Leftrightarrow6x+25=10\)

\(\Leftrightarrow6x=-15\)

\(\Leftrightarrow x=-\dfrac{5}{2}\)

\(c,\left(x+4\right)^2+\left(1-x\right)\left(1+x\right)=7\)

\(\Leftrightarrow x^2+8x+16+1-x^2=7\)

\(\Leftrightarrow8x+17=7\)

\(\Leftrightarrow8x=-10\)

\(\Leftrightarrow x=-\dfrac{5}{4}\)

\(d,\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)

\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)

\(\Leftrightarrow-8x+20=6\)

\(\Leftrightarrow-8x=-14\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

#\(Urushi\)

a, Thay x = 3 và y = -6 vào bt ta đc

\(5.3-4.\left(-6\right)=15-\left(-24\right)=39\\ b,\\ 2.\left(-2\right)^2-5.4=8-20=\left(-12\right)\\ c,\\ 5.\left(-1\right)^2+3.\left(-1\right)-1=5+\left(-3\right)-1=1\)

a) Thay x=3; y=-6

\(5x-4y=5.3-4.\left(-6\right)=15+24=39\)

b) Thay x=-2; y=4

\(2x^4-5y=2.\left(-2\right)^4-5.4=32-20=12\)

c, Thay x=0

\(5x^2+3x-1=5.0+3.0-1=-1\)

+) x=-1

\(5x^2+3x-1=5.\left(-1\right)^2+3.\left(-1\right)-1=5-3-1=1\)

+) \(x=\dfrac{1}{3}\)

\(5x^2+3x-1=5.\left(\dfrac{1}{3}\right)^2+3.\dfrac{1}{3}-1\)

\(=\dfrac{5}{9}+1-1=\dfrac{5}{9}\)

a/

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

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

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

c, \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

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

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

\(=-3x^2+7x-4\)

Thay x = 2 vào A được:

\(=-3.2^2+7.2-4=-2\)

Vậy: Giá trị của A khi x = 2 là -2

==========

b/ \(B=126y^3+\left(x-5y\right)\left(x^2+25y^2+5xy\right)\)

\(=126y^3+x^3-125y^3\)

Thay x = -5 và y = -3 vào B được: 

\(126.\left(-3\right)^3+\left(-5\right)^3-125.\left(-3\right)^3=-152\)

Vậy: Giá trị của B tại x = -5 và y = -3 là -152

==========

c/ \(C=a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(=a^3+b^3-\left(a-b\right)^3\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

Thay a = -4 và b = 4 vào C được:

\(2.4^3+3.\left(-4\right)^2.4-3.\left(-4\right).4^2=512\)

Vậy: Giá trị của C tại a = -4 vào b = 4 là 512

a:Ta có: \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

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

\(=-3x^2+7x-4\)

\(=-3\cdot2^2+7\cdot2-4\)

\(=-12-4+14=-2\)

c: Ta có: \(C=a^3+b^3-\left(a-b\right)\left(a^2-2ab+b^2\right)\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

\(=2\cdot4^3+3\cdot\left(-4\right)^2\cdot4-3\cdot\left(-4\right)\cdot4^2\)

\(=128+192+192=512\)

a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)

=>8x+4-18x+3=2x+1

=>-10x+7=2x+1

=>-12x=-6

hay x=1/2

b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)

=>5x-21=6x

=>-x=21

hay x=-21