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câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
1: \(\Leftrightarrow\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
2: \(\Leftrightarrow7x-21=5x+25\)
=>2x=46
=>x=23
3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)
=>4,5x+2=4x+3
=>x=1
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(\Rightarrow x=\dfrac{-1}{6}\)
b, \(25-\left(5-x\right)=-7\)
\(\Rightarrow\) \(5-x=25-\left(-7\right)\)
\(\Rightarrow5-x=32\)
\(\Rightarrow x=5-32\)
\(\Rightarrow x=-27\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{-5}{7}\)
d, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)
a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(x=\dfrac{-1}{6}\)
b)\(25-\left(5-x\right)=-7\)
\(5-x=25-\left(-7\right)\)
\(5-x=32\)
x= -27
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
d)\(2x\left(x-\dfrac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)
⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)
\(\dfrac{x-3}{5-x}\) = \(\dfrac{5}{7}\) (đk \(x\) ≠ 5)
7.(\(x\) - 3) = 5.(5 - \(x\))
7\(x\) - 21 = 25 - 5\(x\)
7\(x\) + 5\(x\) = 25 + 21
12\(x\) = 46
\(x\) = 46 : 12
\(x\) = \(\dfrac{23}{6}\)
\(\dfrac{\left|x-2\right|}{2}\) = \(\dfrac{\left|2x+3\right|}{3}\)
3.|\(x\) - 2| = |2\(x\) + 3|.2
lập bảng ta có:
Theo bảng trên ta có:
\(x\) < - \(\dfrac{3}{2}\) ⇒ -3\(x\) + 6 = - 4\(x\) - 6
-\(3x\) + 4\(x\) = - 6 - 6
\(x\) = -12
- \(\dfrac{3}{2}\) < \(x\) < 2 ⇒ 3\(x\) - 6 = - 4\(x\) - 6
3\(x\) + 4\(x\) = 6 - 6
7\(x\) = 0
\(x\) = 0
2 < \(x\) ⇒ 3\(x\)- 6 = 4\(x\) + 6
4\(x\) - 3\(x\) = - 6 - 6
\(x\) = -12 (loại)
Vậy \(x\) \(\in\) {-12; 0}