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/x/ = 2,5 => x = 2,5 hoặc x = - 2,5
/x/ = - 1,2 => x thuộc tập hợp rỗng
/x/ + 0,573 = 2
/x/ = 2 - 0,573
/x/ = 1,427
Vậy x = 1,427 hoặc x = - 1,427
/x +1:3/ - 4 = -1
/x + 1 : 3 / = (-1) + ( -4 )
/x +1 : 3 / = -5
x + 1 : 3 = -5 hoặc x + 1 : 3 = 5
x + 1 = (-5) x 3 hoặc x+1 = 5x3
x + 1 = -15 hoặc x+1 = 15
x = (-15) - 1 hoặc x = 15 - 1
x = -16 hoặc x = 14
71-(33+x)=26
33+x=71-26
33+x=45
x=45-33
x=12
Vậy x=12
45-(x-9):2=6
(x-9):2=45-6
(x-9):2=39
x-9=39.2
x-9=78
x=78+9
x=87
Vậy x=87
\(77-\left(33+x\right)=26\\ \Rightarrow33+x=77-26=51\\ \Rightarrow x=51-33=18\\ 45-\left(x-9\right):2=6\\ \Rightarrow\left(x-9\right):2=45-6=39\\ \Rightarrow x-9=39\cdot2=78\\ \Rightarrow x=78+9=87\)
a, 3x - 5 = - 7 - 13
3x - 5 = - 20
3x = - 20 + 5
3x = - 15
x = -5
b,2x-(-3)=7
2x + 3 = 7
2x = 7- 3
2x = 4
x = 2
c, (x-5)(x+6)=0
\(\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
d, \(|x|-10=-3\)
\(|x|=7\)
\(\Rightarrow x=\orbr{\begin{cases}7\\-7\end{cases}}\)
=.= hk tốt!!
11: |2x-3|-1/3=0
=>|2x-3|=1/3
=>\(\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{10}{3}\\2x=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
12: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)
=>\(\left|x+\dfrac{1}{4}\right|=\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{12}\end{matrix}\right.\)
13: \(\left|x-1\right|-2x=\dfrac{1}{2}\)
=>\(\left|x-1\right|=2x+\dfrac{1}{2}\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}-x+1\right)\left(2x+\dfrac{1}{2}+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
14: \(3x-\left|x+15\right|=\dfrac{5}{4}\)
=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16.25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\)
=>\(x=8.125\)
a/ Ta có lx(x+5)l =x
=> x(x+5) = x hoặc x(x+5) = -x
=> x+5 = x:x=1 hoặc x+5 = (-x):x = -1
=> x = 1-5 = -4 hoặc x = -1-5 = -6
Vậy......
mình quên còn 1 TH là x = 0