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=>|0,7-x|=-5,3+11 = 5,7
=> 0,7-x=5,7 hoặc 0,7-x=-5,7
=> x=-5 hoặc x=6,4
k mk nha
|0,7-x|= -5,3 + 11 = 5,6
\(\Rightarrow\orbr{\begin{cases}0,7-x=-5,6\\0,7-x=5,6\end{cases}\Rightarrow\orbr{\begin{cases}x=6,3\\x=4,9\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(C\left(x\right)=\frac{4x-3}{6}-\frac{5-3x}{3}+\frac{1}{3}\)
\(\frac{4x-3}{6}-\frac{5-3x}{3}+\frac{1}{3}=0\)
\(4x-3-2\left(5-3x\right)+2=0\)
\(4x-1-2\left(5-3x\right)=0\)
\(4x-1-10+6x=0\)
\(10x-11=0\)
\(10x=0+11\)
\(10x=11\)
\(x=\frac{11}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(x-\frac{1}{4}\right)^2=\frac{4}{9}=>\left(x-\frac{1}{4}\right)=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
=> \(x=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\)
b, \(\left(x+0,7\right)^3=-8=>\left(x+0,7\right)^3-2^3\)
=> \(x+0,7=-2=>x=-2-0,7=-2,7\)
c \(2^x+2^{x+3}=144=>2^x+2^x.8=144\)
=>\(2^x.\left(8+1\right)=144=>2^x.9=144=>2^x=16\)
=> \(2^x=2^4=>x=4\)
d, \(5-2:\left|x-2\right|=1=>2:\left|x-1\right|=4\)
=> \(\left|x-2\right|=\frac{2}{4}=\frac{1}{2}\)
vậy \(x-2=\frac{1}{2}hoacx-2=-\frac{1}{2}\)
x = \(\frac{2}{5}\) x = \(\frac{3}{2}\)
a )
\(\left(x-\frac{1}{4}\right)^2=\frac{4}{9}\)
\(\left(x-\frac{1}{4}\right)^2=\left(\pm\frac{2}{3}\right)^2\)
\(\Rightarrow x-\frac{1}{4}=\frac{2}{3}\) hoặc \(x-\frac{1}{4}=-\frac{2}{3}\)
\(x=\frac{2}{3}+\frac{1}{4}\) hoặc \(x=-\frac{2}{3}+\frac{1}{4}\)
\(x=\frac{8}{12}+\frac{3}{12}\) hoặc \(x=-\frac{8}{12}+\frac{3}{12}\)
\(x=\frac{11}{12}\) hoặc \(x=-\frac{5}{12}\)
b)
\(\left(x+0,7\right)^3=-8\)
\(\left(x+0,7\right)^3=\left(-2\right)^3\)
\(\Rightarrow x+0,7=-2\)
\(x=-2-0,7\)
\(x=-2,7\)
c)
\(2^x+2^{x+3}=144\)
\(2^x\cdot1+2^x\cdot2^3=144\)
\(2^x\cdot\left(1+2^3\right)=144\)
\(2^x\cdot\left(1+8\right)=144\)
\(2^x\cdot9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\) \(\Rightarrow x=4\)
d)
\(5-2:\left|x-2\right|=2\)
\(5-\left|x-2\right|=2\cdot2\)
\(5-\left|x-2\right|=4\)
\(\left|x-2\right|=5-4\)
\(\left|x-2\right|=1\)
\(\left|x-2\right|=\pm1\)
\(x-2=1\) hoặc \(x-2=-1\)
\(x=1+2\) hoặc \(x=-1+2\)
\(x=3\) hoặc \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
`x-0,7<8`
`=>x<8,7`
Mà `8<x`
`=>8<x<8,7`
Nếu `x in ZZ=>x in` $\varnothing$
`x in R=>8<x<8,7`
Ta có: x-0.7<8<x
\(\Leftrightarrow x< 8.7\)
mà 8<x
nên 8<x<8,7
Vậy: S={x|8<x<8,7}