a) /x+\(\frac{4}{15}\)/ - / -3,75/ = -2,15

=> \(\orbr{\begin{cases}x+\frac{4}{15}+3,75=-2,15\\x+\frac{4}{15}+3,75=2,15\end{cases}}\)

=> ....v.....v giải ra ( từng th )

bài khác tương tự

ai biet tra loi cho mik voi mik dag rat can gap

chi can lam bai c,d,e la duoc

nhung bai con lai mik tu lam duoc

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. \(\dfrac{3}{4}-\left|2x+1\right|=\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{-1}{8}\)

=> \(\left\{{}\begin{matrix}2x+1=\dfrac{-1}{8}\\2x+1=\dfrac{1}{8}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{-9}{16}\\x=\dfrac{-7}{16}\end{matrix}\right.\)

#Yiin

b. \(2.\left|2x-3\right|=\dfrac{1}{2}\)

=> \(\left|2x-3\right|=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)

c: =>3|2x-5|=7,5+4,5=12

=>|2x-5|=4

=>2x-5=4 hoặc 2x-5=-4

=>2x=9 hoặc 2x=1

=>x=9/2 hoặc x=1/2

d: =>|x+4/15|=-2,15+3,75=1,6

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

Gửi em

\(a,\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{2}\\2x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-\dfrac{1}{2}\\2x=-\dfrac{1}{2}-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\2x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)  \(b,\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\\ \Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\\ \Leftrightarrow\left(\dfrac{3+4}{6}\right)x=\dfrac{-10}{3}+\dfrac{3}{3}\\ \Leftrightarrow\dfrac{7}{6}x=\dfrac{-7}{3}\\ \Leftrightarrow x=\left(-\dfrac{7}{3}\right):\dfrac{7}{6}\\ \Leftrightarrow x=-2\)

Vậy \(x=0;x=-\dfrac{1}{2}\)                 Vậy \(x=-2\)

\(c,\dfrac{x-12}{4}=\dfrac{1}{2}\\ \Leftrightarrow2.\left(x-12\right)=4\\ \Leftrightarrow2x-24=4\\ \Leftrightarrow2x=24+2\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=26:2=13\)

Vậy \(x=13\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(\text{#}Toru\)

a) \(\left|x-1,7\right|+\dfrac{1}{2}=23\)

\(\left|x-1,7\right|=22,5\)

\(\Rightarrow x-1,7=\pm22,5\)

\(\Rightarrow x=\left\{24,2;-20,8\right\}\)

b) \(\left|x+\dfrac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)

\(\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\dfrac{4}{15}\right|=1,6\)

\(\Rightarrow x+\dfrac{4}{15}=\pm1,6\)

\(\Rightarrow x=\left\{\dfrac{4}{3};-\dfrac{28}{15}\right\}\)

a)\(\left|x-1,7\right|+\dfrac{1}{2}=23\)

<=>\(\left|x-1,7\right|=22,5\)

<=>\(-\left(x-1,7\right)=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x+1,7=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x=-22,5-1,7\) hoặc \(x=22,5+1,7\)

<=>\(x=24,2\)

vậy \(x=24,2\)