e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)

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

l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)

m) \(=7xy\left(2x-3y+4xy\right)\)

n) \(=2\left(x-y\right)\left(5x-4y\right)\)

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

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

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

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

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)

\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)

\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)

\(\Rightarrow x=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}.\)

c) \(5-\left|3x-1\right|=3\)

\(\Rightarrow\left|3x-1\right|=5-3\)

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

\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)

d) \(\left(1-2x\right)^2=9\)

\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow1-2x=\pm3.\)

\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}.\)

Chúc bạn học tốt!

a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)

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

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

\(=-\left(2x-4\right)\left(x+8\right)\)

b) \(x^3+x^2y-15x-15y\)

\(=x^2\left(x+y\right)-15\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-15\right)\)

c) \(3\left(x+8\right)-x^2-8x\)

\(=3\left(x+8\right)-x\left(x+8\right)\)

\(=\left(x+8\right)\left(3-x\right)\)

d) \(x^3-3x^2+1-3x\)

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

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

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

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

d) \(5x^2-5y^2-20x+20y\)

\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)

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

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

1) \(16x\left(2-x\right)-\left(4x-5\right)^2=0\)

\(32x-16x^2-16x^2+40x-25=0\)

\(72x-16x^2-25=0\)

Đề sai ko bạn

2) \(\left(x-7\right)^2+3=\left(x-2\right)\left(x+2\right)\)

\(\left(x^2-14x+7\right)+3-\left(x-2\right)\left(x+2\right)=0\)

\(x^2-14x+7+3-x^2+4=0\)

\(-14x+14=0\)

\(x=1\)

3) \(\left(2x-3\right)^2-\left(7x-2x\right)^2=2\)

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

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

\(\left(-3x-3\right)\left(7x-3\right)=2\)

=> lập bảng tìm x

4) \(\left(5x-7\right)^2-\left(1-3x\right)^2=16x\left(x-3\right)\)

\(25x^2-70x+49-9x^2+6x-1-16x^2+48x=0\)

\(-16x+48=0\)

\(x=3\)

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

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

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)