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`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}
![](https://rs.olm.vn/images/avt/0.png?1311)
=>|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)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\times\left(x-\frac{1}{7}\right)=0\)
\(x-\frac{1}{7}=0:2=0\)
\(x=0+\frac{1}{7}\)
\(x=\frac{1}{7}\)
vậy \(x=\frac{1}{7}\)
\(\frac{x}{-4}=\frac{-3}{5}\)
\(\Leftrightarrow5x=-3.\left(-4\right)=12\)
\(x=\frac{12}{5}\)
vậy \(x=\frac{12}{5}\)
\(\frac{2,1}{0,7}=\frac{x}{3}\)
\(\Leftrightarrow2,1.3=0,7x\)
\(6,3=0,7x\)
\(x=6,3:0,7\)
\(x=9\)
vậy \(x=9\)
a) 2x(x - 1/7) = 0 b) x/-4 = -3/5 c) 2,1/0,7 = x/3
TH1: 2x = 0 => x = 0 x = -3/5 x 4 2,1 . 3 = 0,7 . x
TH2: x -1/7 = 0 => x = 0 + 1/7 => x = 1/7 x = -12/5 6,3 = 0,7 . x
x = 6,3 : 0, 7
x = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và \(x+z=18\)
Áp dụng t/c dãy tỏ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{4}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=12\end{cases}}\)
b ) \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}\) và \(y-x=39\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{y-x}{-6-5}=\frac{39}{-11}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{39}{-11}\\\frac{y}{-6}=\frac{39}{-11}\\\frac{z}{7}=\frac{39}{-11}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{195}{11}\\y=-\frac{234}{11}\\z=\frac{273}{11}\end{cases}}\)
![](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 \(x=1\)
=> x + 0,7 = -4 => x = -4,7
bn nguyễn hà lan anh đã học về phần nguyên chưa vậy