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HH
24 tháng 6 2018
Giải:
a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)
\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)
\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)
\(\Leftrightarrow30x-37=4\)
\(\Leftrightarrow30x=41\)
\(\Leftrightarrow x=\dfrac{41}{30}\)
Vậy ...
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)
\(\Leftrightarrow x^3+8-x^3-3=14x\)
\(\Leftrightarrow5=14x\)
\(\Leftrightarrow x=\dfrac{5}{14}\)
Vậy ...
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
\(\Leftrightarrow x^3+1-x^3-3x=2\)
\(\Leftrightarrow1-3x=2\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
25 tháng 6 2018
a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)
=> \(x^2-2x-x-3x+7+9x=6\)
=> \(x^2-2x-x^2-3x+7+9x=6\)
=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)
=> \(4x=-1\)
Vậy \(x=\dfrac{-1}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)
=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)
=> \(42x=43\)
Vậy \(x=\dfrac{43}{42}\)
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)
=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)
=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)
=> \(5=14x\)
Vậy \(x=\dfrac{5}{14}\)
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)
=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)
=> \(-3x=1\)
Vậy \(x=\dfrac{-1}{3}\)
17 tháng 9 2023
`x^3 - 3x^2y + x + 3xy^2 - y - y^3`
`=(x)^3 - 3*(x)^2*y + 3*x*y^2 - (y)^3 + (x - y)`
`= (x - y)^3 + (x - y)`
`= (x - y)[(x - y)^2 + 1]`
`= (x - y)(x - y - 1)(x - y + 1)`
____
`@` CT:
`(A - B)^3=A^3-3A^2B+3AB^2- B^3`
16 tháng 10 2016
\(5x\left(x-3\right)=x-3\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)
16 tháng 10 2016
a)\(2x\left(x-2016\right)-2x+4032=0\)
\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)
\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)
b)\(5x\left(x-3\right)=x-3\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)
c)\(\left(3x-1\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-1\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)
3 tháng 11 2019
1. (-2x - 1)(x2 - x - 3) - (x + 2)(x + 1)2
= -2x3 + 2x2 + 6x - x2 + x + 3 - (x + 2)(x2 + 2x + 1)
= -2x3 + x2 + 7x + 3 - x3 - 2x2 - x - 2x2 - 2x - 2
= -3x3 - 3x2 + 4x + 1
2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3
=> (x + 2)(x - 1 - x + 3) = 3
=> (x + 2).0 = 3
...(xem lại đề)
3 tháng 11 2019
\(\left(x+2\right)\left(x-1\right)-\left(x-3\right)\left(x+2\right)=3\)
\(\Leftrightarrow\left(x+2\right)\left(x-1-x+3\right)=3\)
\(\Leftrightarrow2\left(x+2\right)=3\)
\(\Leftrightarrow x+2=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{3}{2}-2\)
\(\Leftrightarrow x=-\frac{1}{2}\)
TC
0
\(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
khỉ nghĩ như này..
x3-3x2=0
(=)x2 (x-3)=0
(=)x2=0,hoac x-3=0
(=)x=3