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30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
CM
24 tháng 9 2018
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
29 tháng 8 2021
Đặt \(x^2+2x=a\), pt trở thành:
\(a^2+2a=15\Leftrightarrow a^2+2a-15=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\\a=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\x^2-2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\x\in\varnothing\left[x^2-2x+5=\left(x-1\right)^2+4>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
6 tháng 8 2021
b)x2-2x+1=4
⇔(x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c)x2-4x+4=9
⇔ (x-2)2=9
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d)4x2-4x+1=4
⇔ (2x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
e)x2-2x-8=0
⇔ x2-4x+2x-8=0
⇔ x(x-4)+2(x-4)=0
⇔(x-4)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
f)9x2-6x-8=0
⇔ 9x2-12x+6x-8=0
⇔ 3x(3x-4)+2(3x-4)=0
⇔ (3x-4)(3x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
8 tháng 12 2023
1: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
2: \(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
3: \(3x-3y+x^2-y^2\)
\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
4: \(5x-5y+x^2-y^2\)
\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(5+x+y\right)\)
5: \(x^2-5x-y^2-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
6: \(x^2-y^2+2x-2y\)
\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+2\right)\)
7: \(x^2-4y^2+x+2y\)
\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
9: \(x^2-4y^2+2x+4y\)
\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+2\right)\)
14 tháng 10 2021
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
14 tháng 10 2021
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
9 tháng 1 2023
a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
13 tháng 11 2021
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(\left(-2x+x^2\right).\left(-2x+x^2\right).\left(-2x+x^2\right).\left(-2x+x^2\right).\left(-2x+x^2\right)=1\)
\(\Leftrightarrow\left(-2x+x^2\right)^5=1\)
\(\Leftrightarrow-2x+x^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x=1-\sqrt{2}\\x=\sqrt{2}+1\end{cases}}\)
Vậy \(x=1-\sqrt{2}\) hoặc \(x=\sqrt{2}+1\)