VD
2
K
Khách
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VD
1
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
LL
4
19 tháng 9 2020
a)
\(-4x\left(-2x+1\right):-4x-\left(x+2\right)=8\)
\(-2x+1-x-2=8\)
\(-3x-1=8\)
\(-3x=9\)
\(x=-3\)
b)
\(-\frac{1}{2}x^2\left(-4x^2+6x-2\right):\left(\frac{-1}{2}x^2\right)+4\left(x^2-2x+1\right)==0\)
\(-4x^2+6x-2+4x^2-8x+4=0\)
\(-2x+2=0\)
\(-2x=-2\)
\(x=1\)
KN
30 tháng 10 2019
a) \(2x^2+3x-8=0\)
Ta có: \(\Delta=3^2+4.2.8=73\)
pt có 2 nghiệm
\(x_1=\frac{-3+\sqrt{73}}{4}\);\(x_1=\frac{-3-\sqrt{73}}{4}\)
d) \(\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
Đặt \(x^2+2x=t\)
\(pt\Leftrightarrow t^2-2t-3=0\)
Ta có: \(\Delta=2^2+4.3=16,\sqrt{\Delta}=4\)
pt trên có 2 nghiệm
\(x_1=\frac{2+4}{2}=3;x_2=\frac{2-4}{2}=-1\)
\(\Rightarrow\orbr{\begin{cases}x^2+2x=3\\x^2+2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{cases}}\)
\(\Rightarrow x\in\left\{-3;-1;1\right\}\)
KN
30 tháng 10 2019
c) \(x^4+8x^3+19x^2+12x=0\)
\(\Leftrightarrow x^4+4x^3+4x^3+16x^2+3x^2+12x=0\)
\(\Leftrightarrow\left(x^4+4x^3+3x^2\right)+\left(4x^3+16x^2+12x\right)=0\)
\(\Leftrightarrow x\left(x^3+4x^2+3x\right)+4\left(x^3+4x^2+3x\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+3x\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^3+x^2+3x^2+3x\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2+3x\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow x\in\left\{0;-1;-3;-4\right\}\)
S
1
KN
22 tháng 12 2019
1. \(x^2-8x+16=\left(x-4\right)^2\)
2. \(\left(x+5\right)\left(x-5\right)=x^2-25\)
3. \(x^3-6x^2+12x-8\)
\(=\left(x^3-8\right)-\left(6x^2-12x\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-4x+4\right)\)
\(=\left(x-2\right)^3\)
4. \(=\left(x+2\right)\left(x^2-2x+4\right)=x^3+8\)
5. \(x^2+2x+1=\left(x+1\right)^2\)
6. \(x^2-1=\left(x+1\right)\left(x-1\right)\)
\(8x^3+\left(x+8\right)^2=8\left(x+2\right)\left(x^2-2x+4\right)\)
\(\Leftrightarrow8x^3+x^2+16x+64=8\left(x^3+8\right)\)
\(\Leftrightarrow8x^3+x^2+16x+64=8x^3+64\)
\(\Leftrightarrow8x^3+x^2+16x+64-8x^3-64=0\)
\(\Leftrightarrow x^2+16x=0\)
\(\Leftrightarrow x\left(x+16\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+16=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-16\end{array}\right.\)
\(8x^3+\left(x+8\right)^2=8\left(x+2\right)\left(x^2-2x+4\right)\)
\(\Leftrightarrow8x^3+\left(x^2+16x+61\right)=8\left(x^3+2^3\right)\)
\(\Leftrightarrow8x^3+x^2+16x+61=8x^3+61\)
\(\Leftrightarrow8x^3+x^2+16x+61-8x^3-61=0\)
\(\Leftrightarrow x^2+16x=0\)
\(\Leftrightarrow x\left(x+16\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+16=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-16\end{array}\right.\)
\(\text{Vậy x=0 hoặc x=-16 }\)