Bài 1a:

\(\frac{4,6}{1,4}\) = \(\frac{11,5}{5x}\)

4,6 x 5\(x\) = 11,5 x 1,4

23\(x\) = 16,1

\(x\) = 16,1 : 23

\(x\) = 0,7

Vậy \(x\) = 0,7

Bài 1b:

\(\frac{x^2}{6}\) = \(\frac{24}{25}\)

25\(x^2\) = 24.6

25\(x^2\) = 144

\(x^2\) = 144/25

\(x\) = - 12/5 hoặc \(x\) = 12/5

Vậy \(x\) ∈ {-12/5; 12/5}

Bài 1c:

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

(\(x\) + 2)(1 - \(x\)) = 1.2

\(x\) - \(x^2\) + 2 - 2\(x\) = 2

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

\(x^2\) + \(x\) + 0 = 0

\(x\)(\(x\) + 1) = 0

\(x\) = 0

\(x\) + 1 = 0

\(x\) = - 1

Vậy \(x\) ∈ {-1; 0}

a) \(\frac{3x-2}{1\frac{2}{5}}=\frac{2\frac{3}{7}}{2\frac{3}{5}}\Rightarrow3x-2=1.\frac{2}{5}.\frac{2\frac{3}{7}}{2\frac{3}{5}}=\frac{7}{5}.\frac{17}{7}.\frac{5}{13}\)

\(3x-2=\frac{17}{13}\Rightarrow3x=2+\frac{17}{13}=\frac{43}{13}\Rightarrow x=\frac{43}{39}.\)

b) \(\frac{x}{0,16}=\frac{9}{x}\Rightarrow x^2=9.0,16=1,44\Rightarrow x=\pm1,2\)

7.(x-1)= 5.(x+5)

7.x-7. = 5.x+25

7.x -5.x =25+7

2.x.    = 32

         x  = 32:2

       x.    = 16

làm rõ ràng, đầy đủ chi tiết luôn

\(\begin{array}{l}a)\frac{x}{{ - 3}} = \frac{7}{{0,75}}\\ \Rightarrow x.0,75 = ( - 3).7\\ \Rightarrow x = \frac{{( - 3).7}}{{0,75}} =  - 28\end{array}\)

Vậy x = 28

\(\begin{array}{l}b) - 0,52:x = \sqrt {1,96} :( - 1,5)\\ - 0,52:x = 1,4:( - 1,5)\\ x = \dfrac{(-0,52).(-1,5)}{1,4}\\x = \frac{39}{{70}}\end{array}\)

Vậy x = \(\frac{39}{{70}}\)

\(\begin{array}{l}c)x:\sqrt 5  = \sqrt 5 :x\\ \Leftrightarrow \frac{x}{{\sqrt 5 }} = \frac{{\sqrt 5 }}{x}\\ \Rightarrow x.x = \sqrt 5 .\sqrt 5 \\ \Leftrightarrow {x^2} = 5\\ \Leftrightarrow \left[ {_{x =  - \sqrt 5 }^{x = \sqrt 5 }} \right.\end{array}\)

Vậy x \( \in \{ \sqrt 5 ; - \sqrt 5 \} \)

Chú ý:

Nếu \({x^2} = a(a > 0)\) thì x = \(\sqrt a \) hoặc x = -\(\sqrt a \)

a: \(\dfrac{x}{-3}=\dfrac{7}{0.75}=\dfrac{28}{3}\)

=>\(x=\dfrac{28\left(-3\right)}{3}=-28\)

b: \(-\dfrac{0.52}{x}=\dfrac{\sqrt{1.96}}{-1.5}=\dfrac{1.4}{-1.5}\)

=>\(x=0.52\cdot\dfrac{1.5}{1.4}=\dfrac{39}{70}\)

c: \(\dfrac{x}{\sqrt{5}}=\dfrac{\sqrt{5}}{x}\)

=>\(x^2=5\)

=>\(x=\pm\sqrt{5}\)

lộn na !!

7x - 7 - 5x - 35 = 0

2x + 28 = 0

2x = -28

x = -14 

Câu a:

\(\frac{x+5}{x+9}\) = \(\frac{x+4}{x+3}\)

(\(x\) + 5)(\(x+3\)) = (\(x+4\))(\(x+9\))

\(x^2\) + 3\(x\) + 5\(x\) + 15 = \(x^2\) + 9\(x\) + 4\(x\) + 36

\(x^2+3x+5x\) - \(x^2-9x\) - 4\(x\) = 36 - 15

(\(x^2-x^2\)) + (\(3x+5x\) - 9\(x-4x\)) = 21

0 + (8\(x\) - 9\(x\) - 4\(x\)) = 21

-\(x-4x\) = 21

-5\(x\) = 21

\(x\) = 21 : (-5)

\(x\) = - \(\frac{21}{5}\)

Vậy \(x=-\frac{21}{5}\)

Bài 1:

a) \(\frac{x}{-15}=\frac{-60}{x}\Rightarrow x^2=\left(-60\right).\left(-15\right)=900\Rightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)

Bài 2: Đặt \(\frac{x}{4}=\frac{y}{7}=k\Rightarrow x=4k;y=7k\)

\(\Rightarrow xy=4k.7k=28k^2=112\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

\(\Rightarrow\orbr{\begin{cases}x=4.2=8\\x=-4.2=-8\end{cases}}\)

Và \(\orbr{\begin{cases}y=7.2=14\\y=-7.2=-14\end{cases}}\)

Bài 3: \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)

\(\Rightarrow\frac{4}{3}:\frac{4}{5}=\frac{2}{3}:\frac{1}{10}x\Rightarrow\frac{5}{3}=\frac{2}{3}:\frac{1}{10}x\)

\(\Rightarrow\frac{1}{10}x=\frac{2}{5}\Rightarrow x=4\)

Mk trả lời nốt bài 4 hộ bn MMS_Hồ Khánh Châu nha:
Bài 4:
Gọi x là giá trị chung của 2 phân số trên.
Ta có: \(\frac{a}{b}=\frac{c}{d}=x\)
\(\Rightarrow a=x.b \)
      \(c=x.d\)
Ta lại có: 
\(\frac{a+c}{b+d}=\frac{x.b+x.d}{b+d}=\frac{x.\left(b+d\right)}{b+d}=x\)
Và \(\frac{a}{b}=x\)
\(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\)
Vậy \(\frac{a}{b}=\frac{a+c}{b+d}\)
Hk tốt nha

Câu a:

0,4 : x = x : 0,9

x^2 = 0,4 x 0,9

x^2 = 0,36

x^2 = (0,6)^2

x = - 0,6 hoặc x = 0,6

Vậy x ∈ {-0,6; 0,6}

Câu b:

13\(\frac13\) : 1\(\frac13\) = 26 : (2x - 1)

\(\frac{40}{13}\) : \(\frac43\) = 26 :(2x - 1)

\(\frac{40}{13}\) x \(\frac34\) = 26 :(2x - 1)

\(\frac{30}{13}\) = 26 : (2x - 1)

26 : \(\frac{30}{13}\) = 2x - 1

26 x \(\frac{13}{30}\) = 2x - 1

169/15 = 2x - 1

2x = 169/15 + 1

2x = 184/15

x = 184/15 : 2

x = 184/15 x 1/2

x = 92/15

Vậy x = 92/15

bài 1

\(\frac{3x-y}{x+y}=\frac{3}{4}\)

\(\Rightarrow\left(3x-y\right).4=\left(x+y\right)3\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)

Vậy \(\frac{x}{y}=\frac{7}{9}\)