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| x + 4/15 | - | - 3,75 | = - | - 2,15 |
| x + 4/15 | - 3,75 = - 2,15
| x + 4/15 | = - 2,15 + 3,75
| x + 4/15 | = 1,6
Trường hợp 1 : x + 4/15 = 1,6 suy ra x = 4/3
Trương hợp 2: x + 4/15 = - 1,6 suy ra x = -28/15
/x+4/15/-/-3,75/=-/-2,15/
/x+4/15/-15/4=-\(\frac{43}{20}\)
x+4/15=\(\frac{43}{20}+\frac{15}{4}\)
x+4/15=59/10
x=59/10-4/15
x=\(5\frac{19}{30}\)
/x + \(\frac{4}{15}\)/ - 3,75 = -2,15
=> /x + \(\frac{4}{15}\)/ = 1,6
=> \(x+\frac{4}{5}=1,6\) hoặc \(x+\frac{4}{5}=-1,6\)
=> \(x=0,8\)hoặc \(x=-2,4\)
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
\(\left|x+\frac{4}{15}\right|=1,6\)
Ta có 2 trường hợp
trường hợp 1
\(x+\frac{4}{15}=1,6\)
\(x=1,6-\frac{4}{15}\)
\(x=\frac{4}{3}\)
Trường hợp 2
\(x+\frac{4}{15}=-1,6\)
\(x=-1,6-\frac{4}{15}\)
\(x=\frac{-28}{15}\)
Vậy \(x=\left\{-\frac{28}{15};\frac{4}{3}\right\}\)
a)|x+4/15|-3,75=-2,15
|x+4/15|=-2,15+3/75
|x+4/15|=1,6
=>x+4/15=1,6 hoặc x+4/15=-1,6
x=1,6-4/15 x=-1,6-4/15
x=4/3 x=-28/15
b)|x-5/3|<1/3
*)x-5/3<1/3
x<1/3+5/3=6/3
x<2
*)x-5/3<-1/3
x<-1/3+5/3
x<4/3
=>x<4/3
Vậy x<4/3
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
=> \(\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}\)
\(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\)
=> \(\left(\frac{1}{2}+\frac{1}{5}\right)x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x=-\frac{3}{5}\)
=> \(x=\left(-\frac{3}{5}\right):\frac{7}{10}=\left(-\frac{3}{5}\right)\cdot\frac{10}{7}=\left(-\frac{3}{1}\right)\cdot\frac{2}{7}=-\frac{6}{7}\)
b) \(\left|2\frac{1}{2}+x\right|-\left(-\frac{2}{3}\right)=3\)
=> \(\left|\frac{5}{2}+x\right|+\frac{2}{3}=3\)
=> \(\left|\frac{5}{2}+x\right|=\frac{7}{3}\)
=> \(\orbr{\begin{cases}\frac{5}{2}+x=\frac{7}{3}\\\frac{5}{2}+x=-\frac{7}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{29}{6}\end{cases}}\)
c) \(\left|x+\frac{4}{15}\right|-\left|-3.75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)
a, \(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow\frac{7}{10}x+\frac{3}{5}=0\Leftrightarrow x=-\frac{6}{7}\)
b, đề sai
c, \(\left|\frac{x+4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\frac{x+4}{15}-3,75=-2,15\Leftrightarrow\frac{x+4}{15}=\frac{8}{5}\Leftrightarrow x+4=24\Leftrightarrow x=28\)
a) | x - 1,5 | = 2
=>x - 1,5 =2 hoặc x - 1,5 = -2
Th1:
x - 1,5 =2
x=2+1,5
x=3,5
Th2:
x - 1,5 = -2
x=-2+1,5
x=-0,5
Vậy x\(\in\){3,5:-0,5}
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Rightarrow\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\Rightarrow\left|x+\frac{4}{15}\right|=-2,15+3,75\)
\(\Rightarrow\left|x+\frac{4}{15}\right|=1,6\)
\(\Rightarrow\left|x+\frac{4}{15}\right|=\frac{8}{5}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-\frac{28}{15}\end{array}\right.\)
Ta có : \(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\frac{4}{15}\right|=3,75-2,15=\frac{8}{5}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{4}{15}=\frac{8}{5}\\-x-\frac{4}{15}=\frac{8}{5}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-\frac{28}{15}\end{array}\right.\)