a: =>10+3x-3=6x+10

=>3x-3=6x

=>-3x=3

=>x=-1

b: =>x+1=0 hoặc x-2=0

=>x=-1 hoặc x=2

a) \(10+3\left(x-1\right)=10+6x\)

\(\Rightarrow10+3x-3=10+6x\)

\(\Rightarrow3x-6x=10-10+3\)

\(\Rightarrow-3x=3\)

\(\Rightarrow x=-\dfrac{3}{3}\)

\(\Rightarrow x=-1\)

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

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

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

\(a,\Leftrightarrow x^2+2x+1-x^2+3x-2x=3\\ \Leftrightarrow3x=2\Leftrightarrow x=\dfrac{3}{2}\\ b,\Leftrightarrow x^2-x-6-x^2+6x-9=15\\ \Leftrightarrow5x=30\Leftrightarrow x=6\\ c,\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=0\\ \Leftrightarrow x=-4\)

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

\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)

b) \(\left(x+2\right)\left(x-3\right)-\left(x-3\right)^2=15\Rightarrow x^2-3x+2x-6-\left(x^2-6x+9\right)=15\)

\(\Rightarrow5x=30\Rightarrow x=6\)

Bài 2:

a: 4x(x-3)+6(3-x)=0

=>4x(x-3)-6(x-3)=0

=>(x-3)(4x-6)=0

=>\(\left[{}\begin{matrix}x-3=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\)

b: \(x^3-x\left(x+1\right)\left(x-1\right)=14\)

=>\(x^3-x\left(x^2-1\right)=14\)

=>\(x^3-x^3+x=14\)

=>x=14

c: \(\left(x^2-x\right)^2+2\left(x^2-x\right)=8\)

=>\(\left(x^2-x\right)^2+2\left(x^2-x\right)-8=0\)

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

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

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

=>\(\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x-2\right)\left(x+1\right)=0\)

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

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

`3(x-1)(x-5) =0`

`<=> (x-1) =0` hoặc `x-5 = 0`.

`<=> x =1` hoặc `x = 5`.

Vậy `x = 1` hoặc `x = 5.`

`b, 3x^2 + 7x = 10`.

`<=> 3x^2 + 7x - 10 = 0`

`<=> (3x+10)(x-1) =0`

`<=> 3x + 10 = 0` hoặc `x - 1=0`

`<=> x = -10/3` hoặc `x = 1.`

Vậy `x = -10/3` hoặc `x = 1.`

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

\(\Leftrightarrow x^2-8x+16-x^2+9=5\)

\(\Leftrightarrow-8x=-20\)

hay \(x=\dfrac{5}{2}\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

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

b) \(\left(x+3\right)^2-5x-15=0\)

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

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

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

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

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)

` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

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

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

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`

a,

 \(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

b,

\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)

c,

a) x - 452 = 77 + 48

x - 452 = 125

x= 125 + 452 

x= 577

b) x + 58 = 64 + 58

x + 58 = 122

x= 122 - 58

x= 64

c) x - 1 - 2 - 3 - 4 = 0

x= 0 + 4 + 3 + 2 +1

x= 10

a)x-452=77+48

x-452=125

x=125+452

x=577

b)x+58=64+58

x+58=122

x=122-58

x=64

c)x-1-2-3-4=0

x=0+4+3+2+1

x=10

Nhớ tick cho mk nha^^

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)

b)\(=>x-1-x-2=0\)

\(=>-3=0\left(vl\right)\) => x ko tồn tại