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a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)
\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)
\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)
\(\dfrac{x}{3}=\dfrac{29}{15}\)
\(x=5,8\)
b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)
\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{7}{2}\)
\(x=\dfrac{7}{4}\)
x^3 = 27
x^3 = 3^3
=> x = 3
(2x-1)^3 = 8
(2x-1)^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3:2
=> ko có x phù hợp.
(x-2)^2 = 16
(x-2)^2 = 4^2
x-2 = 4
x = 4+2
x = 6
Chúc bn học tốt!
x^3 = 27
x^3 = 3^3
Vậy x = 3
Đề bài 2 hình như sai bạn ạ
( x - 2 )^2 = 16
( x- 2 )^2 = 4^2
x - 2 = 4
x = 4 + 2
x = 6
Vậy x = 6
a)
-2x+3=-7
-2x=-7-3
-2x=-10
x=-10:-2
x=5
b)(-3)x+1=-8
-3x=-8-1
-3x=-9
x=-9 :-3
x=3
c)[2x+1]=5 ( cái này à trị tuyệt đối đúng k ?) nếu dấu [ là dấu giá trị tuyệt đối
=>\(\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
=>\(\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
=>\(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy x \(\in\left\{2;-3\right\}\)
d)|2x-1|-3=18
|2x-1|=21
=> \(\orbr{\begin{cases}2x-1=21\\2x-1=-21\end{cases}}\)
=>\(\orbr{\begin{cases}2x=22\\2x=-20\end{cases}}\)
=>\(\orbr{\begin{cases}x=11\\x=-10\end{cases}}\)
vậy \(x\in\left\{11;-10\right\}\)
a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)
\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)
\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)
\(x\cdot\frac{19}{6}=-\frac{1}{24}\)
x = -1/76
b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
2x = 12
x = 6
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
\(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
\(\Leftrightarrow x=\frac{3}{2}\).