Bài 1:

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

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

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

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

\(\Rightarrow9x=7y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}.\)

Vậy \(\dfrac{x}{y}=\dfrac{7}{9}.\)

Bài 1

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

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

\(\Leftrightarrow12x-4y=3x+3y\)

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

\(\Leftrightarrow9x=7y\)

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

Bài 2 : 

Ta có : 3x + 2y = y

=> 3x + y = 0

Lại có ; \(\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-3}{5}=\frac{3x-3}{6}=\frac{3x-3+y+3}{6+1}=\frac{3x+y}{6}=\frac{0}{6}=0\)

Nên \(\frac{x-1}{3}=0\Rightarrow x-1=0\Rightarrow x=1\)

       \(y-3=0\Rightarrow y=3\)

         \(\frac{z-3}{5}=0\Rightarrow z-3=0\Rightarrow z=3\)

Vậy x = 1 , y = 3 , z = 3

1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)

                                               \(\Rightarrow x=30\)

b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)

                               \(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)

                                \(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)

Vậy \(x=\frac{4}{5}\)

2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)

\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)

\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)

\(x=\frac{15}{608}:2=\frac{15}{1216}\)

Vậy \(x=\frac{15}{1216}\)

b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)

\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)

\(\Rightarrow0,25x=\frac{20}{3}.3=20\)

\(\Rightarrow x=20:0,25=80\)

Vậy x = 80

c) \(0,01:2,5=\left(0,75x\right):0,75\)

\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)

\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)

\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)

Vậy \(x=\frac{1}{250}\)

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

\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)

\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)

\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)

Vậy \(x=\frac{100}{9}\)

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

3)  \(\frac{x}{4}=\frac{y}{7}\Leftrightarrow\frac{x^2}{4}=\frac{xy}{7}\Leftrightarrow\frac{x^2}{4}=\frac{112}{7}\Leftrightarrow x^2=\frac{448}{7}=64\Leftrightarrow x=\orbr{\begin{cases}8\\-8\end{cases}}\)                                                             \(\Rightarrow y=\orbr{\begin{cases}112:8=14\\112:\left(-8\right)=-14\end{cases}}\)

Bài 2:

a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)

Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)

\(\Rightarrow4x+12=6x\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\)

Vậy x = 6

b) Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)

\(=\frac{14-5}{8}=\frac{9}{8}\)

+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)

+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)

+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)

Vậy ...

c) \(5^x+5^{x+1}+5^{x+2}=3875\)

\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)

\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)

\(\Rightarrow5^x.31=3875\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

Vậy x = 3

Bài 1 hình như sai đề bạn ạ?

Bài 1: Ta có:  \(\frac{x}{4}=\frac{y}{7}\Rightarrow7x=4y\) (1)

=> 7xy=4yy

=> 7.112=4.y²

=> y²=784:4

=> y²=196.

Mà vì 196= 14.14  => y=14  (2)

TỪ (1) và (2)  => 14.4=x.7

=> x=56:7=8

Vậy x=8;y=14

Bài 1:

a) \(\frac{x-3}{x+5}=\frac{5}{7}\)

\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-5x=25+21\)

\(\Rightarrow2x=46\)

\(\Rightarrow x=46:2\)

\(\Rightarrow x=23\)

Vậy \(x=23.\)

b) \(\frac{7}{x-1}=\frac{x+1}{9}\)

\(\Rightarrow\left(x+1\right).\left(x-1\right)=7.9\)

\(\Rightarrow x^2-x+x-1=63\)

\(\Rightarrow x^2-1=63\)

\(\Rightarrow x^2=63+1\)

\(\Rightarrow x^2=64\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

Vậy \(x\in\left\{8;-8\right\}.\)

c) \(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow x+4=\pm10.\)

\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

Vậy \(x\in\left\{6;-14\right\}.\)

Bài 2:

Ta có: \(\frac{a+5}{a-5}=\frac{b+6}{b-6}.\)

\(\Rightarrow\frac{a+5}{b+6}=\frac{a-5}{b-6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

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

\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{\left(a+5\right)-\left(a-5\right)}{\left(b+6\right)-\left(b-6\right)}=\frac{\left(a-a\right)+\left(5+5\right)}{\left(b-b\right)+\left(6+6\right)}=\frac{10}{12}=\frac{5}{6}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{5}{6}\left(đpcm\right).\)

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