a: \(-3x^2\cdot\left(\dfrac{4}{3}x^2+\dfrac{2}{3}x^2-\dfrac{1}{3}\right)\)

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

b: \(\left(x-3y\right)\left(3x^2+5xy+4y^2\right)\)

\(=3x^3+5x^2y+4xy^2-9x^2y-15xy^2-12y^3\)

\(=3x^3-4x^2y-11xy^2-12y^3\)

c: \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

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

\(=10^2=100\)

d: \(x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^3-16x-x^4+1\)

a: \(=x^5+1-x^5+1=2\)

b: \(=\left(6b^3+2b^2-5b-2\right)\left(3b^2-b+3\right)\)

\(=18b^5-6b^4+18b^3+6b^4-2b^3+6b^2-15b^3+5b^2-15b-6b^2+2b-6\)

\(=18b^5+b^3+5b^2-13b-6\)

c: \(=\left(2a^2+2ab+b^2\right)\cdot2a\left(b^2+2a^2-2ab\right)\)

\(=2a\left[\left(2a^2+b^2\right)^2-4a^2b^2\right]\)

\(=2a\left(4a^4+b^4\right)=8a^5+2ab^4\)

Bài này chỉ cần phá ngoặc là xong 

a, PT <=> \(-x^3-4x+8=15\)

\(x^3+4x+7=0\)( vô nghiệm )

b, PT <=> \(24x+25=49\)

\(x=1\)

a: \(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)

=-65

b \(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)

=27

c: \(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

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

\(=-3x^2+3x-3x+3x^2=0\)

a) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)

b) \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3=\frac{1}{27}x^3+8y^3\)

c) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-\left(3y\right)^3=x^3-27y^3\)

d) \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{2}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{9}\)

( x + 4 )( x² - 4x + 16 ) = x³ + 4³ = x³ + 64

( 1/3x + 2y )( 1/9x² - 2/3xy + 4y² ) = ( 1/3x )³ - ( 2y )³ = 1/27x³ - 8y³

( x - 3y )( x² + 3xy + 9y² ) = x³ - ( 3y )³ = x³ - 27y³

( x² - 1/3 )( x⁴ + 1/3x² + 1/9 ) = ( x² )³ - ( 1/3 )³ = x⁶ - 1/27

HĐT số 6 + 7 bạn nhé ^^

\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)

\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)

bạn có thể làm câu D,E được không ạ

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

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

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

b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)

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

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

\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)