x+2/x+1+3/x-2=3/x^2-x-2+1 giúp mk với
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
Y
23 tháng 6 2023
\(1,\left(3x+2\right)\left(5-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)
\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)
\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{9}{46}\)
Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)
\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)
\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)
\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)
\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)
\(\Leftrightarrow6x-4=16\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
Vậy \(S=\left\{\dfrac{10}{3}\right\}\)
\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)
\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Rightarrow5x-5=4x+8\)
\(\Rightarrow x=13\left(tmdk\right)\)
Vậy \(S=\left\{13\right\}\)
4 tháng 11 2018
a) \(\left(x+1\right)^2-2\left(x+1\right)\left(3-x\right)+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+2\left(x+1\right)\left(x-3\right)+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+1+x-3\right)^2=0\)
\(\Leftrightarrow\left(2x-2\right)^2=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
Vậy x = 1
b) \(\left(x+2\right)^2-2\left(x+2\right)\left(x-8\right)+\left(x-8\right)^2=0\)
\(\Leftrightarrow\left(x+2-x+8\right)^2=0\)
\(\Leftrightarrow\)\(\left(0x+10\right)^2=0\)
=> Phương trình vô nghiệm
19 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
19 tháng 10 2021
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
1 tháng 9 2023
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
1 tháng 9 2023
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
22 tháng 6 2021
`1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1)`
ĐK:`x ne 1`
`pt<=>(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x(x-1))/(x^3-1)`
`<=>x^2+x+1-3x^2=2x^2-2x`
`<=>4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+x-1=0`
`<=>(x-1)(4x+1)=0`
`x ne 1=>x-1 ne 0`
`<=>4x+1=0`
`<=>x=-1/4`
Vậy `S={-1/4}`
NN
1
24 tháng 11 2023
(2x - 1) ( x+ 7 ) ; ( x - 2 ) ( x + 3) là 2 câu riêng ạ?
hay đều gộp lại ạ?
22 tháng 6 2021
đk: x khác 2;4
\(\dfrac{2}{\left(x-4\right)\left(2-x\right)}-\dfrac{x-1}{x-2}-\dfrac{x+3}{x-4}=0\)
<=> \(\dfrac{-2-\left(x-1\right)\left(x-4\right)-\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)
<=> \(-2-x^2+5x-4-x^2-x+6=0\)
<=> -2x2 +4x = 0
<=> 2x(x-2) = 0
Mà x khác 2
<=> 2x = 0
<=> x = 0
\(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
\(\Leftrightarrow3+x^2-x-2=x^2-4+3x+3\)
\(\Leftrightarrow x^2-x+1=x^2+3x-1\)
=>-4x=-2
hay x=1/2(nhận)
thank you