Cứu với ạ
8x3+1:2x+1=
X3+x2-12:x-2= x3 +x2 +2x +2:x2+2
số sau x là mũ ạ
K
Khách
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Những câu hỏi liên quan
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
KK
13 tháng 9 2021
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
30 tháng 9 2021
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
VT
2
TA
17 tháng 7 2021
a) `(x^3-x^2)/(x^3-2x^2+x)`
`=(x^2(x-1))/(x(x-1)(x-1))`
`=x/(x-1)`
`=>` 2 phân thức bằng nhau.
b) `(x^2+2x+1)/(2x^2-2)`
`=((x+1)(x+1))/(2(x+1)(x-1))`
`=(x+1)/(2(x-1))`
`=(x+1)/(2x-2)`
`=>` 2 phân thức bằng nhau
17 tháng 7 2021
a) Ta có: \(\dfrac{x^3-x^2}{x^3-2x^2+x}\)
\(=\dfrac{x^2\left(x-1\right)}{x\left(x^2-2x+1\right)}\)
\(=\dfrac{x\cdot\left(x-1\right)}{\left(x-1\right)^2}=\dfrac{x}{x-1}\)
b) Ta có: \(\dfrac{x^2+2x+1}{2x^2-2}\)
\(=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{2x-2}\)
28 tháng 1 2022
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
9 tháng 11 2021
\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)
(8x3+1):(2x+1)=((2x)3+1):(2x+1)=(2x+1)(4x2−2x+1):(2x+1)=4x2−2x+1
x2 + 3x + 6
(\(x^3\) + \(x^2\) - 12) : (\(x\) - 2)
= [(\(x^3-8)\) + (\(x^2\) - 4)] :(\(x-2\))
= [(\(x-2\))(\(x^2+2x+4)\) + \(\left(x-2\right)\left(x+2\right)\)] :(\(x-2\))
= (\(x-2\))(\(x^2+2x+4\) + \(x+2\)):(\(x-2\)
= (\(x-2):\left(x-2\right)\).[\(x^2\) + (2\(x\) + \(x\)) + (4 + 2)]
= 1.[\(x^2\) + 3\(x\) + 6]
= \(x^2+3x+6\)