tìm x ,biết
a :( x-7 )^x+1 - ( x-7 )^x+1 = 0
b : | x + 4/5 | - | 3,75| = - | 2,15 |
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a, 11/13 - ( 5/42 - x ) = - (5/28 - 11/13)
11/13 - (5/42 - x) = - 5/28 + 11/13
- (5/42 - x) + 5/28 = -11/13 + 11/13
- 5/42 + x + 5/28 = 0
- 5/42 + x = 0 - 5/28
- 5/42 + x = - 5/28
x = -5/28 +5/42
x = - 5/84
b, / 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
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
Vậy x = 4/3 hoặc x = - 28/15
c, ( 0,25 - 30% x ) . 1/3 = 1/4 - 31/6
( 1/4 - 3/10 x ) . 1/3 = - 59/12
( 1/4 - 3/10 x ) = - 59/12 : 1/3
1/4 - 3/10 x = - 59/4
3/10 x = 1/4 + 59/4
3/10 x = 15
x = 15 : 3/10
x = 50
d, ( x - 1/2 ) : 1/3 + 5/7 = 68/7
( x - 1/2 ) : 1/3 = 68/7 - 5/7
( x - 1/2 ) : 1/3 = 63/7
( x - 1/2 ) = 63/7 . 1/3
x -1/2 = 3
x = 3 + 1/2
x = 7/2
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
\(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=-\dfrac{8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\\ \Rightarrow\left|x+\dfrac{4}{15}\right|=1,6=\dfrac{8}{5}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=-\dfrac{8}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
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
\(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\)
=> \(\left(\frac{1}{2}+\frac{1}{5}\right)x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x=-\frac{3}{5}\)
=> \(x=\left(-\frac{3}{5}\right):\frac{7}{10}=\left(-\frac{3}{5}\right)\cdot\frac{10}{7}=\left(-\frac{3}{1}\right)\cdot\frac{2}{7}=-\frac{6}{7}\)
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=7 b, x=.....