\(3x\left(x^2+2y\right)^2-12xy\left(x^2+y\right)\)

\(=3x\left(x^4+4x^2y+4y^2\right)-12xy\left(x^2+y\right)\)

\(=3x^5+12x^3y+12xy^2-12xy\left(x^2-y\right)\)

\(=3x^5+\left(12x^3y\right)^2-\left(12xy^2\right)^2\)

\(=3x^5\)

Ta có : 

\(\left|1-2x\right|-\left|3x+1\right|=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}1-2x=3x+1\\1-2x=-3x-1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x+2x=1-1\\-2x+3x=-1-1\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x=0\\x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy \(x=0\) hoặc \(x=-2\)

Chúc bạn học tốt ~ 

cảm ơn Phùng Minh Quân nhiều !!!

u^2v^2(u+v)^2-(u^2v+uv^2)^2 - Step-by-Step Calculator - Symbolab

Tham khảo ở đó nhé!

bn có thể tham khảo mà đúng ko 

a)để  \(2x^2-4x\)dương

\(\Leftrightarrow2x^2-4x>0\)

\(\Leftrightarrow2x\left(x-2\right)>0\)

TH1: \(\Rightarrow\hept{\begin{cases}2x>0\\x-2>0\end{cases}\Rightarrow\orbr{\begin{cases}x>0\\x>2\end{cases}}\Rightarrow x>2}\)

TH2: \(\hept{\begin{cases}2x< 0\\x-2< 0\end{cases}\Rightarrow\orbr{\begin{cases}x< 0\\x< 2\end{cases}}\Rightarrow x< 0}\)

Nhân hết ra là xong, ez ghê

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=10\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=10\)

\(\Leftrightarrow18x-2=10\Leftrightarrow x=\frac{2}{3}\)

\(27y^3+9y^2+y+\frac{1}{27}\)

\(=\left(3y+\frac{1}{3}\right)^3\)

p/s: chúc bạn học tốt

bạn giải chi tiết ra hộ mk với

\(27y^3+9y^2+y+\frac{1}{27}\)

\(=\left(3y\right)^3+3\cdot\left(3y\right)^2\cdot\frac{1}{3}+3\cdot3y\cdot\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3\)

\(=\left(3y+\frac{1}{3}\right)^3\)

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

DỂ QUÁ!!!!!!!!!!!!!!!!!!!!!!!!

tui hk biết làm

\(4x^2y^4-\left(3xy^2-1\right)^2\)

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

\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)

p/s: chúc bạn học tốt