Phân tích thành nhân tử :
a) x3 - 4x2 + 8x - 8
b) (2xy + 1)2 - ( 2x + y)2
c) 1 + 6x - 6x2 - x3
tìm x :
a) x2 - 9 - 2(x+3)2 = 0
b) (5x + 1)2 = (2x - 3)2
K
Khách
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NV
1
29 tháng 7 2019
\(2xy-x^2+3y^2-4y+1\)
\(=-\left(x^2-2xy+y^2\right)+4y^2-4y+1\)
\(=-\left(x-y\right)^2+\left(2y-1\right)^2\)
\(=\left(2y-1+x-y\right)\left(2y-1-x+y\right)\)
\(=\left(y+x-1\right)\left(3y-x-1\right)\)
NV
2
B
29 tháng 7 2019
\(2x^2-5x+2=2x^2-4x-x+2=2x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(2x-1\right)\)
KN
29 tháng 7 2019
a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)
14 tháng 10 2020
\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
a) ĐK : x ≠ ±3
\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b) Để A < 2
=> \(\frac{3x}{x-3}< 2\)
<=> \(\frac{3x}{x-3}-2< 0\)
<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)
<=> \(\frac{3x-2x+6}{x-3}< 0\)
<=> \(\frac{x+6}{x-3}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)
2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )
Vậy -6 < x < 3
BN
0
PT
2
HH
1
29 tháng 7 2019
\(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\)
\(=x^4+4x^2+4-x^4+16=4x^2+20=4\left(x^2+5\right)\)
KN
29 tháng 7 2019
\(\left(x^2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=x^4-\left(x^4-16\right)\)
\(=x^4-x^4+16=16\)
HL
29 tháng 7 2019
\(=\left(x^2\right)^2-\left(x^2-14\right)\left(x^2+4\right)\)
\(=x^4-x^4+16\)
\(=16\)
\(\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(5x+1+2x-3\right)\left(5x+1-2x+3\right)=0\)
\(\Rightarrow\left(7x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\3x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=\frac{-4}{3}\end{cases}}}\)
Vậy.......
\(a,x^3-4x^2+8x-8\)
\(=\left(x^3-8\right)-\left(4x^2-8x\right)\)
\(=\left(x^3-2^3\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4-x+2\right)\)
\(=\left(x-2\right)\left(x^2+3x+6\right)\)
\(b,\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\cdot\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+1\cdot\left(y+1\right)\right]\cdot\left[2x\left(y-1\right)-1\cdot\left(y-1\right)\right]\)
\(=\left[\left(y+1\right)\left(2x+1\right)\right]\cdot\left[\left(y-1\right)\left(2x-1\right)\right]\)
\(=\left(y+1\right)\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(=\left(y^2-1\right)\left(4x^2-1\right)\)
\(c,1+6x-6x^2-x^3\)
\(=\left(1-x^3\right)-\left(6x^2-6x\right)\)
\(=\left(1^3-x^3\right)-6x\left(x-1\right)\)
\(=\left(1-x\right)\left(1+2x+x^2\right)+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+2x+x^2+6x\right)\)
\(=\left(1-x\right)\left(1+8x+x^2\right)\)
\(a,x^2-9-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x^2-3^2\right)-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2\cdot\left(x+3\right)\right]=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2x-6\right]=0\)
\(\Rightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\-x-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\-x=9\Rightarrow x=-9\end{cases}}\)
\(b,\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2\div\left(2x-3\right)^2=0\)
\(\Rightarrow\left[\left(5x+1\right)\div\left(2x-3\right)\right]^2=0\)
\(\Rightarrow\left(5x+1\right)\div\left(2x-3\right)=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow x=-\frac{1}{5}\)