\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\left(x< 0\right)\)

\(D=x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)

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

\(=x-4+\left|x-4\right|\)

\(=x-4+x-4\)

\(=2x-8\)

\(C=\sqrt{9x^2}-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{x^2-8x+16}=x-4+x-4=2x-8\)

a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)

\(=64-x^3+\left(x-4\right)^3\)

\(=64-x^3+x^3-12x^2+48x-64\)

\(=-12x^2+48x\)

b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(=27x^3+8-27x^3+8\)

=16

c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)

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

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

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

b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

=>-6x+16=0

=>-6x=-16

hay x=8/3(nhận)

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

\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)

\(\Leftrightarrow2x^2+4x-2x^2+2=0\)

=>4x+2=0

hay x=-1/2(nhận)

\(a,=\left(x+3\right)^3=\left(-3+3\right)^3=0\\ b,=27x^3+1-\left(1-27x^3\right)=27x^3+1-1+27x^3=54x^3\\ =54\cdot10^3=54\cdot1000=54000\)

c, hình như sai đề á e

where is đề

a) 20x-5y=5(4x-y)

b) 5x(x-1)-3x(x-1)=(5x-3x)(x-1)=2x(x-1)

c) x(x+y)-6x-6y=x(x+y)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)

d) 6x³-9x²=3x²(2x-3)

e) 4x²y - 8xy² + 10x²y²=2xy(2x-4y+5xy)

g) 20x²y-12x³=4x²(5y-3x)

h) 8x⁴+ 12x²y⁴- 16x³y⁴ = 4x²(2x²+3y⁴-4xy⁴)

k) 4xy² + 8xyz=4xy(y+2z)

l) 3x(x + 1) - 5y( x + 1 )=(x+1)(3x-5y)

m) 4x²-1=(2x-1)(2x+1)

o) 9-(x-y)²=(3-x+y)(3+x-y)

p) x³+27=(x+3)(x²+3x+9)

n) (x-y)²-4=(x-y-2)(x-y+2)

r) x⁴ + 2x² + 1=(x⁴+x²)+(x²+1)=x²(x²+1)+(x²+1)=(x²+1)²

s) 4x²  - 12xy + 9y²  = (2x-3y)²

t) x² - x - y² - y =(x²-y²)-(x+y)=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

v) x³ - x + y³ - y = (x³+y³)-(x+y)=(x+y)(x²-xy+y²)-(x+y)=(x+y)(x²-xy+y²-1)

a: \(20x-5y=5\left(4x-y\right)\)

b: \(5x\left(x-1\right)-3x\left(x-1\right)=2x\left(x-1\right)\)

c: \(x\left(x+y\right)-6x-6y=\left(x+y\right)\left(x-6\right)\)

d: \(6x^3-9x^2=3x^2\left(x-3\right)\)

e: \(4x^2y-8xy^2+10x^2y^2=2xy\left(2x-4y+5xy\right)\)

a) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow\left(3x-2\right)\left[\left(3x\right)^2+3x\cdot2+2^2\right]-\left(3x-1\right)\left[\left(3x\right)^2+3x\cdot1+1\right]=x-4\)

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

\(\Leftrightarrow x=-3\) ( thỏa mãn )

P/s : Đề câu b) viết lại nhé, mình không hiểu lắm :))

\(9\left(2x+1\right)=4\left(x-5\right)^2\)

\(\Leftrightarrow18x+9=4\left(x^2-10x+25\right)\)

\(\Leftrightarrow18x+9=4x^2-40x+100\)

\(\Leftrightarrow4x^2-58x+91=0\)

Ta có \(\Delta=58^2-4.4.91=1908,\sqrt{\Delta}=6\sqrt{53}\)

\(\Rightarrow x=\frac{58\pm6\sqrt{53}}{8}\)

a: \(81x^5-x^3\)

\(=x^3\left(81x^2-1\right)\)

\(=x^3\left(9x-1\right)\left(9x+1\right)\)

b: \(9x^2y-12xy+4y\)

\(=y\left(9x^2-12x+4\right)\)

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

c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)

\(=\left(x-5\right)^2-\left(4x-8\right)^2\)

\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)

\(=-3\left(x-1\right)\left(5x-13\right)\)

d: Ta có: \(9x^2-y^2-21x-7y\)

\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)

\(=\left(3x+y\right)\left(3x-y-7\right)\)

e: Ta có: \(-y^2+8y-16+9x^2\)

\(=-\left(y^2-8y+16-9x^2\right)\)

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

f: Ta có: \(5x^2-4x-1\)

\(=5x^2-5x+x-1\)

\(=5x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left(5x+1\right)\)

c: \(x^2+4x+4=\left(x+2\right)^2\)

d: \(9x^2+6x+1=\left(3x+1\right)^2\)

(a) \(9x^2+12x+4=0\)

\(\Leftrightarrow\left(3x+2\right)^2=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\dfrac{3}{2}\)

(b) \(x^2+\dfrac{1}{4}=x\)

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

(c) \(4-\dfrac{12}{x}+\dfrac{9}{x^2}=0\left(x\ne0\right)\)

\(\Leftrightarrow\left(2-\dfrac{3}{x}\right)^2=0\Leftrightarrow2-\dfrac{3}{x}=0\Leftrightarrow x=\dfrac{3}{2}\)