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1/ -3x(x+2)^2+(x+3)(x-1)(x+1)-(2x-3)^2
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21 tháng 2 2021
\(\dfrac{1}{3}x^2y^5\left(\dfrac{-3}{5}x^3y\right)+x^5y^6=\dfrac{-1}{5}x^5y^6+x^5y^6=\dfrac{4}{5}x^5y^6\)
21 tháng 2 2021
1: Ta có: \(\dfrac{1}{3}x^2y^5\cdot\left(-\dfrac{3}{5}x^3y\right)+x^5y^6\)
\(=\dfrac{-1}{5}x^5y^6+x^5y^6\)
\(=\dfrac{4}{5}x^5y^6\)
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
7 tháng 8 2017
1) 2x.(5x-3x)+2x.(3x-5)-3.(x-7)=3
10x-6x^2+6x^2-10x-3x+21=3
-3x =-18
suy ra x=6
2) 3x.(x+1) -2x.(x+2)=-1-x
3x^2 +3x-2x^2-4x =-1-x
x^2 =-1
suy ra không có giá trị nào của x thỏa mãn đề bài
3) 2x^2 +3.(x^2-1)=5x(x+1)
2x^2 +3x^2-3 =5x^2+5x
-5x =3
x=-3/5
giải rồi đấy
nhớ tích đúng nha :)
4 tháng 3 2020
a) (x - 2)3 + (3x - 1)(3x + 1) = (x + 1)3
<=> x3 - 6x2 + 12x - 8 + 9x2 - 1 - x3 - 3x2 - 3x - 1 = 0
<=> 9x - 10 = 0
<=> 9x = 10
<=> x = 10/9
Vậy S = {10/9}
b) (x + 1)(2x - 3) = (2x - 1)(x + 5)
<=> 2x2 - x - 3 - 2x2 - 9x + 5 = 0
<=> -10x + 2 = 0
<=> -10x = -2
<=> x = 1/5
Vậy S = {1/5}
c) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> -5x2 + 2x + 5x2 + x + 22 - 1 = 0
<=> 3x = -21
<=> x = -7
Vậy S = {-7}
d) (x - 3)(x + 4) - 2(3x - 2) = (x - 4)2
<=> x2 + x - 12 - 6x + 4 - x2 + 8x - 16 = 0
<=> 3x - 24 = 0
<=> 3x = 24
<=> x = 8
Vậy S = {8}
e) x(x + 3)2 - 3x = (x + 2)3 + 1
<=> x3 + 6x2 + 9x - 3x = x3 + 6x2 + 12x + 8 + 1
<=> x3 + 6x2 + 6x - x3 - 6x2 - 12x = 9
<=> -6x = 9
<=> x = -3/2
Vậy S = {-3/2}
f) (x + 1)(x2 - x + 1) - 2x = x(x + 1)(x- 1)
<=> x3 + 1 - 2x = x3 - x
<=> x3 - 2x - x3 + x = -1
<=> -x = -1
<=> x = 1
Vậy S = {1}
-3x( x + 2 )2 + ( x + 3 )( x - 1 )( x + 1 ) - ( 2x - 3 )2
= -3x( x2 + 4x + 4 ) + ( x + 3 )( x2 - 1 ) - ( 4x2 - 12x + 9 )
= -3x3 - 12x2 - 12x + x3 - x + 3x2 - 3 - 4x2 + 12x - 9
= ( -3x3 + x3 ) + ( -12x2 + 3x2 - 4x2 ) + ( -12x - x + 12x ) + ( -3 - 9 )
= -2x3 - 13x2 - x - 12