a,  0,4 : x = x : 0,9

<=> x² = 0,4 . 0,9

<=> x² = 0,36

<=> x = 0,6 hoặc -0,6

b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow10=26\div\left(2x-1\right)\)

\(\Leftrightarrow2x-1=\frac{13}{5}\)

\(\Leftrightarrow2x=\frac{18}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\)

c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow6x+7=4\)

\(\Leftrightarrow6x=-3\)

\(\Leftrightarrow x=\frac{-1}{2}\)

d, \(\frac{37-x}{x+13}=\frac{3}{7}\) 

\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)

\(\Leftrightarrow259-7x=3x+39\)

\(\Leftrightarrow-10x=-220\)

\(\Leftrightarrow x=22\)

a, x=0,6

b,x=1,8

c,x=-0,5

d,x=\(\frac{39}{256}\)

1.

a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)

<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26

<=> 10 + 26 = 13x

<=> 13x = 36

<=> x = \(\frac{36}{13}\)

b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)

<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)

<=> x = \(\frac{1}{7}\)

c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)

<=> (37 - x) . 7 = 3.(x + 13)

<=> 119 - 7x = 3x + 39

<=> -7x - 3x = 39 - 119

<=> -10x = -80

<=> x = 8

d) \(\frac{x-1}{x+5}=\frac{6}{7}\)

<=> 7(x - 1) = 6(x + 5)

<=> 7x - 7 = 6x + 30

<=> 7x - 6x = 30 + 7

<=> x = 37

e)

2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)

<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)

<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)

Bài 2. đề sai

Bài 3.

a) 6,88 : x = \(\frac{12}{27}\)

<=> x = 6,88 : \(\frac{12}{27}\)

<=> x = 15,48

b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x

<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x

<=> \(\frac{5}{7}=13:2x\)

<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)

<=> x = 9,1

a) 0,4:x=x:0,9

0,4.0,9=x²

0,36 =x²

\(\sqrt{0,36}\)=x

0,6 =x

a) 0,4 : x = x : 0,9 x^2=0,9 . 0,4 x^2=(0,6)^2 =>x=+_6

a) Đặt \(x-1=a\)

\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)

Vậy pt vô nghiệm

a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2}=2\)

=> không có x thỏa mãn đề bài.

b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)

\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)

\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)

\(7-4x-3x^2=25x-25\)

\(7-4x-3x^2-25x+25=0\)

\(32-29x-3x^2=0\)

\(3x^2+29x-30=0\)

\(3x^2+32x-3x-32=0\)

\(x\left(3x+32\right)-\left(3x+32\right)=0\)

\(\left(3x+32\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)

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