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MT
20 tháng 6 2015
1)5(x^2-1)+x(1-5x)= x-2
<=>5x2-5+x-5x2=x-2
<=>-5+x=x-2
<=>x-x=-2+5
<=>0x=3(vô lí)
vậy ko tìm được x
HH
0
17 tháng 7 2020
Câu 1:
a) Ta có: \(VT=x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)\)=VP(đpcm)
c) Ta có: \(VT=a\left(b+1\right)+b\left(a+1\right)\)
\(=ab+a+ab+b\)
\(=a+b+2ab\)(1)
Thay ab=1 vào biểu thức (1), ta được:
a+b+2(*)
Ta có: VP=(a+1)(b+1)=ab+a+b+1(2)
Thay ab=1 vào biểu thức (2), ta được:
1+a+b+1=a+b+2(**)
Từ (*) và (**) ta được VT=VP(đpcm)
Câu 2:
Ta có: \(\left(x-3\right)\left(x+x^2\right)+2\left(x-5\right)\left(x+1\right)-x^3=12\)
\(\Leftrightarrow x^2+x^3-3x-3x^2+2\left(x^2+x-5x-5\right)-x^3=12\)
\(\Leftrightarrow x^3-2x^2-3x+2x^2-8x-10-x^3-12=0\)
\(\Leftrightarrow-11x-22=0\)
\(\Leftrightarrow-11x=22\)
hay x=-2
Vậy: x=-2
QA
1 tháng 8 2021
Trả lời:
Bài 1:
a, ( x - y ) - 3x ( y - x ) = ( x - y ) + 3x ( x - y ) = ( x - y ) ( 1 + 3x )
b, ( 1 - 2x ) + y ( 2x - 1 ) = ( 1 - 2x ) - y ( 1 - 2x ) = ( 1 - 2x ) ( 1 - y )
Bài 2:
A = x ( x - 1 ) + ( x + y ) ( y - x ) = x2 - x + y2 - x2 = y2 - x
B = ( x - 2 ) ( x + 2 ) - ( x - 1 )2 + 5 = x2 - 4 - x2 + 2x - 1 + 5 = 2x
Bài 3:
a, ( x - 2 ) ( x + 1 ) - ( x - 3 )2 = 0
<=> x2 + x - 2x - 2 - ( x2 - 6x + 9 ) = 0
<=> x2 - x - 2 - x2 + 6x - 9 = 0
<=> 5x - 11 = 0
<=> 5x = 11
<=> x = 11/5
Vậy x = 11/5 là nghiệm của pt.
b, ( x - 1 )2 - 2 ( 1 - x ) = 0
<=> x2 - 2x + 1 - 2 + 2x = 0
<=> x2 - 1 = 0
<=> ( x - 1 ) ( x + 1 ) = 0
<=> x - 1 = 0 hoặc x + 1 = 0
<=> x = 1 hoặc x = - 1
Vậy x = 1; x = - 1 là nghiệm của pt.
PT
17 tháng 10 2017
1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)
\(=\dfrac{1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1+\left(-x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x^2-1}\)
2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}+\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+\left(-x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{-1}{x^2-1}\)
3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{1}{y\left(x-y\right)}+\dfrac{-1}{x\left(x-y\right)}\)
\(=\dfrac{1x}{y\left(x-y\right)x}+\dfrac{-1y}{x\left(x-y\right)y}\)
\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)
\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)
4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)
\(=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}-\dfrac{1x}{\left(x-1\right)x}\)
\(=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{-x}{x\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)-x}{x\left(x-1\right)}\)
\(=\dfrac{-1}{x\left(x-1\right)}\)
5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(=\dfrac{1\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1x}{\left(x+1\right)x}\)
\(=\dfrac{x+1}{x\left(x+1\right)}+\dfrac{-x}{x\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)
6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)
\(=\dfrac{1}{2x\left(x-5\right)}+\dfrac{-2x}{2x\left(x-5\right)}\)
\(=\dfrac{1-2x}{2x\left(x-5\right)}\)
7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x+3\right)}\)
8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)
\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)
\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)
\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)
\(x\left(x-1\right)-y\left(x-1\right)=\left(x-y\right)\left(x-1\right)\)