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a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)

\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)

\(\Leftrightarrow-6x-3=10\)

=>-6x=13

hay x=-13/6

b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)

=>3x-2=-5x-8

=>8x=-6

hay x=-3/4

c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)

=>x-27=20

hay x=47

22 tháng 5 2021

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

22 tháng 5 2021

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

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

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

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

\(< =>2-x=0< =>x=2\)

28 tháng 6 2017

Phép cộng các phân thức đại số

3 tháng 8 2018

|x-2|<3

|x+4|<5

|3+x|>2

really your name is crazy

and do you crazy

21 tháng 7 2021

1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

23 tháng 4 2019

a. \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left[3\left(2x-1\right)-5\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)

b. \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

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

\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)

\(\Leftrightarrow-\left(3x+1\right)\left(x+2\right)=0\)

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

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

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

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

\(\Leftrightarrow3x\left(x+2\right)=0\)

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

d. \(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow\left(2x^3-8x^2\right)+\left(5x^2-20x\right)-\left(12x-48\right)=0\)

\(\Leftrightarrow2x^2\left(x-4\right)+5x\left(x-4\right)-12\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x^2+5x-12\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\)

\(\Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\frac{3}{2}\end{matrix}\right.\)

e. \(x^2+2x-15=0\)

\(\Leftrightarrow\left(x^2-3x\right)+\left(5x-15\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

a: \(x^2\left(2x-3\right)+8x-12=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x^2+4\right)=0\)

=>2x-3=0

hay x=3/2

b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)

=>(2x-5)(x+11)=0

=>x=5/2 hoặc x=-11

c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)

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

hay \(x\in\left\{0;4;-4\right\}\)