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3 tháng 3 2020
a) (2x + 5)(x - 3) = (x - 4)(3 - x)
<=> (2x + 5)(x - 3) + (x - 3)(x - 4) = 0
<=> (2x + 5 + x - 4)(x - 3) = 0
<=> (3x + 1)(x - 3) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\frac{1}{3}\\x=3\end{matrix}\right.\)
Vậy S = {-1/3; 3}
b) 18x2(x + 4) - 12(x2 + 4x) = 0
<=> 18x2(x + 4) - 12x(x + 4) = 0
<=> 6x(x + 4)(3x - 2) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+4=0\\3x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {0; -2; 2/3}
G
1
31 tháng 12 2021
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
4 tháng 10 2018
\(9x^5-18x^4-16x+32=0\)
\(\left(9x^5-18x^4\right)-\left(16x-32\right)=0\)
\(9x^4\left(x-2\right)-16\left(x-2\right)=0\)
\(\left(x-2\right)\left(9x^4-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\9x^4-16=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\9x^4=16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x^4=\frac{16}{9}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\\left(x^2\right)^2=\left(\frac{\pm4}{3}\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\pm\sqrt{\frac{4}{3}}\end{cases}}\)
Vậy,..........
14 tháng 12 2021
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
1 tháng 2
a: \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b: \(x^3-x^2-x-2\)
\(=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\cdot\left(x^2+x+1\right)\)
c: \(x^3+x^2-x+2\)
\(=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
d: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
e: Sửa đề: \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
f: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
g: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
h: \(\left(x^2-3\right)^2+16\)
\(=x^4-6x^2+9+16\)
\(=x^4-6x^2+25\)
\(=x^4+10x^2+25-16x^2\)
\(=\left(x^2+5\right)^2-\left(4x\right)^2\)
\(=\left(x^2+5+4x\right)\left(x^2+5-4x\right)\)
23 tháng 6 2023
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
`2x^3-16x^2+18x=0`
`<=>2x(x^2-8x+9)=0`
`<=>2x(x^2-8x+16-7)=0`
`<=>2x[(x-4)^2-7]=0`
`<=>` $\left[\begin{matrix} 2x=0\\ (x-4)^2-7=0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=0\\ (x-4)^2=7\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=0\\ x-4=\pm \sqrt{7}\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=0\\ x=4 \pm \sqrt{7}\end{matrix}\right.$
Vậy `S={0;4+-\sqrt{7}}`