(x^2-1)^3-(x^4+x^2+1)(x^2-1)
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N
10 tháng 7 2017
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
DP
11 tháng 7 2017
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
T
13 tháng 9 2018
a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)
\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)
\(\Leftrightarrow79x-35=44\)
\(\Leftrightarrow79x=44+35\)
\(\Leftrightarrow79x=79\)
\(\Leftrightarrow x=1\)
b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)
\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)
\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)
\(\Leftrightarrow17x+10=27\)
\(\Leftrightarrow17x=17\)
\(\Leftrightarrow x=1\)
DK
1
15 tháng 4 2020
\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x-9=0
<=> -2x=9
<=> \(x=\frac{-9}{2}\left(tmđk\right)\)
27 tháng 9 2021
a: \(\left(x-1\right)^3+27\)
\(=\left(x-1+3\right)\left(x^2-2x+1+3x-3+3\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
b: \(\left(x-2\right)^3-8\)
\(=\left(x-2-2\right)\left(x^2-4x+4+2x-4+4\right)\)
\(=\left(x-4\right)\left(x^2-2x+4\right)\)
TK
1
19 tháng 9 2018
=> 1 - 3 . X=x - 7 hoặc 1 - 3 . X =-(x-7)
*1 - 3x =x - 7 *1 - 3x = -(x - 7 )
8 =x + 3x 1 - 3x = -x + 7
8 =4x -3x+x =7-1
8 : 4 =x -2x =6
2 = x x = 6:(-2)
=>x = 2 x = -3
vậy x \(\in\){2; -3}
đúng + x =1
x =1 -đúng
x = thích
(x2 _ 1)3 _ (x4 + x2 + 1) (x2 _ 1)
= x6 _ 1 _ ( x6 + x4 + x2 _ x4 _ x2 _ 1)
= x6 _ 1 _ x6 _ x4 _ x2 + x4 + x2 + 1
= 0