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14 tháng 8 2020
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
NN
15 tháng 11 2017
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NN
15 tháng 11 2017
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
6 tháng 8 2020
Rút gọn
a) Ta có: \(2x^2\left(5x^3-4x^2y-7xy+1\right)\)
\(=10x^5-8x^4y-14x^3y+2x^2\)
b) Ta có: \(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)\cdot x\)
\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
\(=10x^4-7x^3-5x^2\)
2x( 3x3 - x ) - 4x2( x - x2 + 1 ) + ( x - 3x2 )x
= 6x4 - 2x2 - 4x3 + 4x4 - 4x2 + x2 - 3x3
= 10x4 - 7x3 - 5x2
\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
\(=10x^4-5x^2-7x^3\)