a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....

b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........

c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......

d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......

a) \(x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)

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

\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)

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

\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)

c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)

hic, mk chx học

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

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

c)

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

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)