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NL
1
18 tháng 8 2023
a: (1-2x)^3-(1+2x)^3
\(=1^3-3\cdot1^2\cdot2x+3\cdot1\cdot\left(2x\right)^2-8x^3-8x^3-12x^2-6x-1\)
\(=1-6x+12x^2-8x^3-8x^3-12x^2-6x-1\)
\(=-16x^3-12x\)
b: \(=x^3-6x^2+12x-8-x^3-x^2+8\)
\(=-7x^2+12x\)
c: \(=x^3+8-12x+6x^2-x^3+6x^2+12x\)
\(=12x^2+8\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
TP
1
17 tháng 9 2021
1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)
2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)
3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)
4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)
TL
1
18 tháng 12 2022
3: =>x(x+1)=0
=>x=0 hoặc x=-1
4: =>(2x-3)(x+2)=0
=>x=3/2 hoặc x=-2
6: =>6x=7 hoặc 6x=-7
=>x=7/6 hoặc x==7/6
LD
20 tháng 1 2022
\(1,\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow x-2x=-4-7+4\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
\(2,\Leftrightarrow x-1-2x+1=9-x\)
\(\Leftrightarrow x+x-2x=9-1+1\)
\(\Leftrightarrow0x=9\)
\(\Rightarrow x\in\varnothing\)
Vậy \(S=\left\{\varnothing\right\}\)
\(3,\Leftrightarrow2x^2+3x-2x+3=2x^2+10x-x-5\)
\(\Leftrightarrow2x^2-2x^2+3x-2x-10x+x=-5-3\)
\(-8x=-8\)
\(\Rightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(x^3+\left(x-1\right)^3=\left(x+x-1\right)^3-3\left(x+x-1\right)x\left(x-1\right)\)
\(=\left(2x-1\right)^3-3\left(2x-1\right)x\left(x-1\right)\)
Do đó: \(x^3+\left(x-1\right)^3=\left(2x-1\right)^3\)
<=> \(\left(2x-1\right)^3-3\left(2x-1\right)x\left(x-1\right)=\left(2x-1\right)^3\)
<=> (2x - 1) x (x -1 ) = 0
<=> 2x - 1 = 0 hoặc x = 0 hoặc x - 1 = 0
<=> x = 1/2 hoặc x = 0 hoặc x = 1
Kết luận: ...
Bài làm
x3 + ( x - 1 )3 = ( 2x - 1 )3
<=> x3 + x3 - 3x2 + 3x - 1 = 8x3 - 24x2 + 6x - 1
<=> x3 + x3 - 8x3 - 3x2 + 24x2 + 3x - 6x - 1 + 1 = 0
<=> -6x3 + 21x2 - 3x = 0
<=> -x( 6x2 - 21x + 3 ) = 0