Ta có: \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+7x^2-6x=0\)

\(\Leftrightarrow x^2+7x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

bạn có thể tách rõ hơn đoạn cuối dc khum mình cảm ơn

à k cần nữa ạ

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)

\(\left(x-20\right).\left(7x+10\right)=0\)
\(=>\left[{}\begin{matrix}x-20=0\\7x+10=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)

7x(x-20)+10(x-20)=0

(x-20)x(7x+10)=0

(x-20)=0 hoặc (7x+10)=0

x=20 hoặc 7x=-10

x=20 hoặc x=-10/7

a: \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{2x^2+7x+6}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(2+x\right)}{2x^2+4x+3x+6}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(x+2\right)}{\left(x+2\right)\left(2x+3\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{2-x}{2x+3}=\dfrac{2-\left(-2\right)}{2\cdot\left(-2\right)+3}=\dfrac{4}{-4+3}=-4\)

b: \(\lim\limits_{x\rightarrow4}\dfrac{2x^2-13x+20}{x^3+64}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x^2-8x-5x+20}{\left(x+4\right)\left(x^2-4x+16\right)}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(2x-5\right)}{x^3+64}\)

\(=\dfrac{\left(4-4\right)\left(2\cdot4-5\right)}{4^3+64}=0\)

c: \(\lim\limits_{x\rightarrow-1}\dfrac{2x^2+8x+6}{-2x^2+7x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x^2+2x+6x+6}{-2x^2-2x+9x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{-2x\left(x+1\right)+9\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{\left(x+1\right)\left(-2x+9\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x+6}{-2x+9}=\dfrac{2\cdot\left(-1\right)+6}{-2\cdot\left(-1\right)+9}\)

\(=\dfrac{4}{11}\)

1)7x(x-5)-x(x-5)=(x-5)(7x-x)=6x(x-5)

2)x⁴+3x³+x+3=x³(x+3)+(x+3)=(x+3)(x³+1)=(x+3)(x+1)(x²-x+1)

3)x⁴+64=[(x²)²+2.x².8+64]-16x²=(x²+8)²-(4x)²=(x²+4x+8)(x²-4x+8)

\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)

`1/3x + 2/3(x-1) =0`

` 1/3x + 2/3x -2/3 = 0`

` ( 1/3 + 2/3) x -2/3 = 0`

` 3/3x -2/3 = 0`

` 1x-2/3 = 0`

`1/x = 0 + 2/3`

` 1x = 2/3`

` x = 2/3`

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

2:

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4

 bài 2 :

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4