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11 tháng 12 2020
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
HO
19 tháng 8 2021
a)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
N
0
28 tháng 5 2017
Đặt \(A=x^{13}-\left(8x^{12}-8x^{11}+8x^{10}-8x^9+.....+8x^2-8x^1\right)+8\)
Đặt \(B=8x^{12}-8x^{11}+8x^{10}-....+8x^2-8x^1\)
\(B=8.\left(x^{12}-x^{11}+x^{10}-x^9+....+x^2-x^1\right)\)
Đặt \(C=x^{12}-x^{11}+x^{10}-x^9+...+x^2-x\)
Suy ra \(C.x=x^{13}-x^{12}+x^{11}-x^{10}+.....+x^3-x^2\)
Nên \(C.x-C=x^{13}-x\)hay \(C.\left(x-1\right)=x^{13}-x\)
Khi đó \(C=\frac{x^{13}-x}{x-1}\)nên\(B=8.\frac{x^{13}-x}{x-1}\)
Từ đó tính tương tự nha , cách làm thì có thể sai những em vẫn cố gắng giúp , ai có cách hay hơn thì giải nhé
30 tháng 12 2019
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow\left(8x^2-8x^2\right)+\left(-24x+8\right)=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=12:\left(-24\right)\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
30 tháng 12 2019
\(8x\left(x-2\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-\left(8x^2-8\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow-24x+8=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=-\frac{1}{2}\)
OP
4 tháng 8 2016
\(8x^3+\left(x+8\right)^2=8\left(x+2\right)\left(x^2-2x+4\right)\)
\(8x^3+x^2+2\times x\times8+8^2=8\left(x^3+2^3\right)\)
\(8x^3+x^2+16x+64+8x^2=8\left(x^3+8\right)\)
\(8x^3+x\times\left(x+16\right)+64=8x^3+64\)
\(8x^3-8x^3+64-64+x\times\left(x+16\right)=0\)
\(x\times\left(x+16\right)=0\)
TH1:
\(x=0\)
TH2:
\(x+16=0\)
\(x=-16\)
Vậy x = 0 hoặc x = -16
\(x\left(x^2+x+1\right)=8\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x-8\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\end{matrix}\right.\)