e) \(\left(x+3\right)^3=\left(2x\right)^3\)

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

\(\Rightarrow2x-x=3\)

\(\Rightarrow x=3\)

f) \(\left(5-x\right)^5=32\)

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

\(\Rightarrow5-x=2\)

\(\Rightarrow x=5-2\)

\(\Rightarrow x=3\)

g) \(\left(5x-6\right)^3=64\)

\(\Rightarrow\left(5x-6\right)^3=4^3\)

\(\Rightarrow5x-6=4\)

\(\Rightarrow5x=4+6\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=\dfrac{10}{2}\)

\(\Rightarrow x=5\)

h) \(5\cdot9^x=405\)

\(\Rightarrow9^x=\dfrac{405}{5}\)

\(\Rightarrow9^x=81\)

\(\Rightarrow9^x=9^2\)

\(\Rightarrow x=2\)

i) \(11^5:11^{n-2}=11^5\)

\(\Rightarrow11^{n-2}=11^5:11^5\)

\(\Rightarrow11^{n-2}=1\)

\(\Rightarrow11^{n-2}=11^0\)

\(\Rightarrow n-2=0\)

\(\Rightarrow n=2\)

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

\(\Rightarrow3x=2x+1\)

\(\Rightarrow3x-2x=1\)

\(\Rightarrow x=1\)

Thay \(x = 2\) vào phương trình \(\sqrt { - 2{x^2} - 2x + 11}  = \sqrt { - {x^2} + 3} \) ta thấy không thỏa mãn vì dưới dấu căn là \( - 1\) không thỏa mãn

Vậy \(x =  2\) không là nghiệm của phương trình do đó lời giải như trên là sai.

a: \(3\left(x-3\right)-6x=0\)

=>\(3x-9-6x=0\)

=>-3x-9=0

=>3x+9=0

=>3x=-9

=>\(x=-\dfrac{9}{3}=-3\)

b: Đề thiếu vế phải rồi bạn

c: \(2\left(x-3\right)+3x=9\)

=>2x-6+3x=9

=>5x-6=9

=>5x=6+9=15

=>x=15/5=3

d: \(x\left(x-11\right)+2\left(x-11\right)=0\)

=>\(\left(x-11\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)

e: \(x\left(x+2\right)+8=x^2\)

=>\(x^2+2x+8=x^2\)

=>2x+8=0

=>2x=-8

=>x=-8/2=-4

f: \(8\left(x+1\right)+2x=-2\)

=>\(8x+8+2x=-2\)

=>10x=-2-8=-10

=>\(x=-\dfrac{10}{10}=-1\)

g: 12-3(x+2)=0

=>3(x+2)=12

=>x+2=12/3=4

=>x=4-2=2

đợi xíu

Giúp mình với mình cần gấp lắm

\(\dfrac{2x-3}{2}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow3\left(2x-3\right)>8x-11\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(2x-3\le8x-11\)

\(\Leftrightarrow-6x\le-8\)

\(\Leftrightarrow x\ge\dfrac{8}{6}\)

Vậy \(S=\left\{x|x\ge\dfrac{8}{6}\right\}\)

\(\dfrac{x-3}{2}>\dfrac{x-11}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{6}>\dfrac{2\left(x-11\right)}{6}\)

\(\Leftrightarrow3\left(x-3\right)>2\left(x-11\right)\)

\(\Leftrightarrow3x-9>2x-22\)

\(\Leftrightarrow x>-13\)

Vậy \(S=\left\{x|x>-13\right\}\)

\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)

\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)

\(\text{⇔}8x+5=-11\) 

\(\text{⇔}8x=-16\)

\(\text{⇔}x=-2\)

Vậy: \(x=-2\)

==========

\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)

\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x=-\dfrac{4}{3}\)

\(\text{⇔}x=\dfrac{2}{39}\)

1)

x^3 -16x=0`

`<=>x(x^2 -16)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b)

`x^4 -2x^3=0`

`<=>x^3 (x-2)=0`

\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

3)

`(2x-11)(x^2 -1)=0`

\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)

4)

`x^3 -36x=0`

`<=>x(x^2 -36)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

5)

`2x+19=0`

`<=>2x=-19`

`<=>x=-19/2`

bài về nghiệm của đa thức

\(-x+3=2x+9\\ \Rightarrow-x-2x=9-3\\ \Rightarrow-3x=6\\ \Rightarrow x=-\dfrac{6}{3}\\ \Rightarrow x=-2\)

Vậy \(x=-2\)

\(-x-5=-2x+11\\ \Rightarrow-x+2x=11+5\\ \Rightarrow x=16\)

Vậy \(x=16\)

b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)

\(=2x^2-6x-2x^2+4x+2x-4\)

=-4