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a,2,5-x=1,3
x=2,5-1,3
x=1,2
b,x-3,5=7,5
x=7.5+3,5
x=11
c,(x-1,5)-2,5=0
x-1,5=2,5
x=2,5+1,5
x=4
d,1,6-(x-0,2)=0
x-0,2=1,6
x=1,6+0,2
x=1,8
A=(3,7- -x)+2,5
vì 2,5>=0 với mọi x
=> x=-1,2
B= (x+1,5)-4,5
Vì -4,5<=0 với mọi x
=> x=-6
Học Tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
7,5 - 3 |5 - 2x| = -4,5 => -3 |5 - 2x| = -12 => |5 - 2x| = 4
=> 5 - 2x = 4 hoặc -4
=> -2x = -1 hoặc -9
=> x = 0,5 hoặc x = 4,5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}:\frac{2}{5}\)
\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{35}{12}\)
\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}\)
\(\Rightarrow x=\frac{35}{4}\)
Vậy \(x=\frac{35}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài : 5
a) Ta có : A = 3 + |4 - x|
Vì : \(\left|4-x\right|\ge0\forall x\)
Nên : A = 3 + |4 - x| \(\ge3\forall x\)
Vậy Amin = 3 khi x = 4
b) Ta có : B = 5|1 - 4x| - 1
Vì \(\text{5|1 - 4x|}\ge0\forall x\)
Nên : B = 5|1 - 4x| - 1 \(\ge-1\forall x\)
Vậy Bmin = -1 khi x = 1/4
a)\(\left|2x-3\right|=6\)
\(\Rightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}...\\...\end{cases}}\)
b)\(2.\left|3x+1\right|=5\)
\(\left|3x+1\right|=2,5\)
\(\Rightarrow\orbr{\begin{cases}3x+1=2,5\\3x+1=-2,5\end{cases}}\Rightarrow\orbr{\begin{cases}...\\...\end{cases}}\)
c)\(7,5-3\left|5-2x\right|=-4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=4\)
\(...\)
![](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)
a)
\(\left(\frac{x-1}{4}\right)^2=\frac{4}{9}\)
⇒ \(\frac{x-1}{4}=\frac{2}{3}\)
\(\Rightarrow\frac{\left(x-1\right).3}{12}=\frac{8}{12}\)
=> (x - 1) = 8/3
=> x = 8/3 - 1
=> x = 5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) \(|x-3,5|=7,5\)
\(\Rightarrow x-3,5=7,5\)
hay \(x-3,5=-7,5\)
TH1 : \(x-3,5=7,5\Rightarrow x=7,5+3,5=11\)
TH2 : \(x-3,5=-7,5\Rightarrow x=-7,5+3,5=-4\)
b) \(|x+\dfrac{4}{5}|-\dfrac{1}{2}=0\)
\(\Rightarrow\left(x+\dfrac{4}{5}\right)-\dfrac{1}{2}=0\) (chỉ có 1 TH vì số 0 ko phải dương or âm)
\(\left(x+\dfrac{4}{5}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{4}{5}=\dfrac{5-8}{10}=\dfrac{-3}{10}\)
c) \(3,6-|x-0,4|=0\)
\(\Rightarrow3,6-\left(x-0,4\right)=0\) ( giải thích giống câu b )
\(\Rightarrow-\left(x-0,4\right)=0-3,6\)
\(\Rightarrow-\left(x-0,4\right)=-3,6\)
\(\Rightarrow-x+0,4=-3,6\) ( Phá dấu )
\(\Rightarrow-x=-3,6-0,4=-3,6+\left(-0,4\right)=-4\)
\(\Rightarrow x=4\)
d) \(-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
\(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
hay \(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{5}{9}\)
TH1 : \(-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{-5}{9}\Rightarrow\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\left(-\dfrac{5}{9}\right)\)
\(\Rightarrow\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{12}.\dfrac{9}{5}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\dfrac{3}{4}=-\dfrac{5.4}{6.3}=-\dfrac{5.2}{3.3}=-\dfrac{10}{9}\)
TH2 : \(\Rightarrow-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{9}\)
\(\Rightarrow\)\(\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\dfrac{5}{9}=-\dfrac{5.9}{12.5}=-\dfrac{9}{12}=-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{4}{3}=\dfrac{10}{9}\)
Vậy x = ....
e)
Vì \(|x-3,5|\ge0;|4,5-x|\ge0\) với mọi x
Do đó : \(|x-3,5|+|4,5-x|=0\)
\(\Rightarrow|x-3,5|=0;|4,5-x|=0\)
\(\Rightarrow x-3,5=0\) và \(4,5-x=0\)
\(\Rightarrow x=0+3,5=3,5\) và \(-x=0+4,5=4,5\Rightarrow x=-4,5\)
( không đồng thời xảy ra)
\(\Rightarrow\) Không tồn tại x thuộc Q để \(|x-3,5|+|4,5-x|=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3/5-2x/=7,5-(-4,5) \(\Rightarrow\)5-2x=4 hoặc 5-2x=-4 Vậy x=0.5 hoặc 4.5
\(\Rightarrow\)3/5-2x/=7,5+4,5 \(\Rightarrow\) 2x=5-4 hoặc2x=5-(-4) Nhớ tích nha!!!!!
\(\Rightarrow\)3/5-2x/=12 \(\Rightarrow\)2x=1 hoặc 2x=9
\(\Rightarrow\)/5-2x/=12:3 \(\Rightarrow\)x=1:2 hoặc x=9:2
\(\Rightarrow\)/5-2x/=4 \(\Rightarrow\)x=0.5 hoặc x=4.5
\(3\left|5-2x\right|=7,5+4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=12\div3=4\)
\(\Rightarrow5-2x=4\)hoặc \(5-2x=-4\)
\(5-2x=4\Rightarrow2x=5-4=1\Rightarrow x=\frac{1}{2}\)
\(5-2x=-4\Rightarrow2x=5+4=9\Rightarrow x=\frac{9}{2}=4,5\)
`a) x *4,5 = 7,5`
`=> x = 7,5 : 4,5`
`=> x = 5/3`
`b) 2/5 - x + 1/2 = 4/9`
`=> 2/5 - x = 4/9 - 1/2`
`=> 2/5 - x = -1/18`
`=> x = 2/5 + 1/18`
`=> x = 41/90`
`c) x - 0,7 = 1,3`
`=> x = 1,3 + 0,7`
`=> x = 2`