7 - I 2x + 1 I = x
Giúp em với ạa
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22 tháng 6 2021
`(x+2)/(x-2)-2/(x^2-2x)=1/x`
ĐK:`x ne 0,x ne +-2`
Nhân 2 vế với `x^2-2x ne 0` ta có pt
`x(x+2)-2=x(x-2)`
`<=>x^2+2x-2=x^2-2x`
`<=>4x=2`
`<=>x=1/2.(tm)`
Vậy `S={1/2}`
25 tháng 2 2022
1: Thay \(x=\dfrac{4-\sqrt{7}}{2}\) vào B, ta được:
\(B=\dfrac{1}{\sqrt{\dfrac{4-\sqrt{7}}{2}}+1}=1:\left(\dfrac{\sqrt{7}-1+2}{2}\right)=1\cdot\dfrac{2}{\sqrt{7}+1}=\dfrac{-1+\sqrt{7}}{3}\)
7 tháng 8 2021
a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
b) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
PL
giúp em với ạa, bài này là bài tập không phải thi đâu ạa
0
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
SV
Sinh Viên NEU
CTVVIP
31 tháng 12 2024
1 Kai is tired because he stayed up late watching TV
2 Since I have a broken leg, I fell over while I was playing basketball
3 Sehun is going to be late for school as the bus is late
4 Because Lisa was careless, she broke the cup
5 Rose wants to go home since she feels sick
SV
Sinh Viên NEU
CTVVIP
31 tháng 12 2024
6 Jimin is hungry as he hasn't eaten all day
7 Since plastic bags are very hard to dissolve, they will cause pollution
8 People reuse and recycle bottles and cans as they want to reduce garbage
9 The sea is becoming increasingly polluted since people drop garbage into the sea
10 We shouldn't throw trash onto the water as polluted water can directly do harm to people's health and kill fish
\(7-\left|2x+1\right|=x\)
\(\Rightarrow\left|2x+1\right|=7-x\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=-7+x\\2x+1=7-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=-7-1\\2x+x=7-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-8\\3x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
=>|2x+1|=7-x
=>x<=7 và (2x+1)^2=(7-x)^2=(x-7)^2
=>x<=7 và (2x+1-x+7)(2x+1+x-7)=0
=>x<=7 và (x+8)(3x-6)=0
=>x=2 hoặc x=-8