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\(\frac{x-1}{6}=\frac{6}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)
a) \(\frac{x-1}{6}=\frac{6}{x-1}\)
<=> (x - 1)(x - 1) = 6.6
<=> (x - 1)2 = 36
<=> (x - 1)2 = 62
\(\Leftrightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)
b) \(\frac{\left|x+3\right|}{5}=\frac{1}{4}\)
\(\Leftrightarrow\frac{5.\left|x+3\right|}{5}=\frac{5.1}{4}\)
\(\Leftrightarrow\left|x+3\right|=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=\frac{5}{4}\\x+3=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{4}\\x=-\frac{17}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{4}\\x=-\frac{17}{4}\end{cases}}\)

a) \(\frac{7}{11}\)
b)\(\frac{218}{165}\)
c)\(\frac{1}{4}\)
d)\(\frac{-21}{20}\)
e)\(\frac{2}{9}\)
g) 1
h)\(\frac{-561}{26}\)
i)\(\frac{4}{3}\)

\(\frac{2}{5}\)= \(\frac{x}{10}\)
2 x 10 = \(x\)x 5
20 = \(x\)x 5
x = 20 : 5
x = 4
Vậy x = 4
Tk mk nha !!
Áp dụng \(\frac{a}{b}=\frac{c}{d}\)\(\Leftrightarrow\)\(ad=bc\) \(\left(a,b,c,d\inℤ;b,d\ne0\right)\) là okkk :)
Ta có :
\(\frac{2}{5}=\frac{x}{10}\)
\(\Leftrightarrow\)\(2.10=5x\)
\(\Leftrightarrow\)\(20=5x\)
\(\Leftrightarrow\)\(x=\frac{20}{5}\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=4\)
Chúc bạn học tốt ~

Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)

1.
a) (—7/3)3:(—7/3)2=(—7/3)3–2=—7/3
b) (—4/9):(—4/9)3= (—4/9)1–3=(—4/9)—2=81/16
c) (1/5)10:(1/5)7=(1/5)10–7=(1/5)3=1/125
2.
a) —x/7 =1/—21
==> —x.(—21)=7.1
==> —x.(—21)=7
==> —x=7:(—21)
==> —x=—1/3
==> x=1/3
b) 4 2/5 . 0,5–1 3/7= 22/5 . 1/2 —10/7= 22.1/5.2–10/7= 11/5 —10/7= 77/35 — 50/35= 27/35
c) 3x2–2x=0
==> x3(3–2)=0
x3.1=0
x3=0:1
x3=0
==> x=0
c) 9x2–1=0
9x2=0+1
9x2=1
x2=1:9
x2=1/9
x2=12/32 hoặc x2=(—1/3)2
Vậy x=1/3 hoặc x=—1/3
\(-\frac57-\left(-\frac{3}{21}\right)\)
\(=-\frac57+\frac{3}{21}\)
\(=-\frac{15+3}{21}\)
\(=-\frac{12}{21}=-\frac47\)
\(\frac{5}{-7}-\frac{-3}{21}\)
\(=\frac{-5}{7}-\frac{-3}{21}\)
\(=\frac{-105}{147}-\frac{-21}{147}\)
\(=\frac{-105-\left(-21\right)}{147}\)
\(=\frac{-105+21}{147}\)
\(=\frac{-84}{147}\)
\(=\frac{-4}{7}\)