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16 tháng 9 2021
(2x -1 ) (x+1 ) - 2x^2 -2x^2 0
Suy ra 2x^2 + 2x - x -1 -2x^2 =0
Suy ra x-1 = 0
Suy ra x = 1
10 tháng 7 2017
Theo đề bài ta có :
\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)
=> \(\frac{\left(3x-x^2\right)}{x+1}\cdot\frac{\left(3-x+x^2+x\right)}{x+1}=2\)
=> \(\left(3x-x^2\right)\left(x^2+3\right)=2\left(x+1\right)^2\)
=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)
=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)
=> \(3x^3+5x-5x^2-x^4-2=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)
=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)
=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)
=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)
=> \(\left(1-x\right)\left(x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right)=0\)
=> \(\left(1-x\right)\left(x-1\right)\left(x^2-x+2\right)=0\)
Ta Thấy :
\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)
=> x = 1
TN
27 tháng 7 2017
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
7 tháng 6 2016
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
7 tháng 6 2016
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
PN
21 tháng 11 2015
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\left(n\in N\right)\)
Như vậy,
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
MT
21 tháng 11 2015
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}=\frac{x+6}{x.\left(x+6\right)}-\frac{x}{x.\left(x+6\right)}=\frac{6}{x^2+6x}\)
4 tháng 6 2016
Ta có: \(B=\frac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\frac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}-x\right)\right]\)
\(=\frac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(1+x+x^2+x\right)\left(1-x+x^2-x\right)\right]\)
\(=\frac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(1+x\right)^2\left(1-x\right)^2\right]\)\(=\frac{x\left(1-x\right)^2\left(1+x\right)^2}{1+x^2}.\frac{1}{\left(1+x\right)^2\left(1-x\right)^2}\)
\(=\frac{x}{1+x^2}\)
29 tháng 7 2016
\(3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2=2x^3-\frac{3}{2}x^2+2\)
\(2x^2-10x-3x-2x^2=26\)
-13x=26
x=-2
TL:
= \(x^3-1\)\(-x^3-6\)
= -7
x= -7
T.I.C.K. Đúng cho mk với