\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}=\frac{27}{9}=3\)

\(\Rightarrow x=\frac{3\cdot4+2}{2}=7\)

\(\Rightarrow y=\frac{3\cdot9+6}{3}=11\)

\(\Rightarrow z=3\cdot4+3=15\)

       \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3x-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}\)

= \(\frac{27}{9}\)

=  \(3\)

\(\Rightarrow\)\(x=\frac{3.4+2}{2}=7\)

\(\Rightarrow\)\(y=\frac{3.9+6}{3}=11\)

\(\Rightarrow\)\(z=3.4+3=15\)

\(\frac{x}{2}=\frac{y}{5}=\frac{2}{3}\Rightarrow\frac{2x}{4}=\frac{3y}{15}=\frac{2}{3}\)

Áp dụng dãy tỉ số = nhau ta có:

\(\frac{2x}{4}=\frac{3y}{15}=\frac{2}{3}=\frac{2x+3y-2}{4+15-3}=\frac{32}{16}=2\)

Suy ra: \(\frac{x}{2}=2\Rightarrow x=4\)

            \(\frac{y}{5}=2\Rightarrow y=10\)

\(\frac{32-x}{7}=\frac{x-42}{9}\)

=\(\frac{\left(32-x\right)9}{63}=\frac{\left(x-42\right)7}{63}\)

\(\Rightarrow\)\(\left(32-x\right)9=\left(x-42\right)7\)

=\(288-x9=x7-294\)

=\(288+294=x9+x7\)

=\(x=-36\frac{6}{16}\)

=\(x\times16=-582\)

\(x=-582\div16\)

a,\(\frac{32-x}{7}=\frac{x-42}{9}\)

\(\Leftrightarrow9\left(32-x\right)=7\left(x-42\right)\)

\(\Leftrightarrow288-9x-7x-294=0\)

\(\Leftrightarrow9x+7x=288-294\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

b. \(\left(2x-1\right)^2+\left|x+3\right|=0\)

\(\Leftrightarrow\left|x+3\right|=-4x^2+4x-1\)

\(\left|x+3\right|=x+3\)khi \(x+3\ge0\)hay \(x\ge-3\)

\(\left|x+3\right|=-\left(x+3\right)\)khi \(x+3< 0\)hay \(x< -3\)

với \(x\ge-3\Rightarrow x+3=-4x^2+4x-1\)

\(\Leftrightarrow4x^2-4x+1+x+3=0\)

\(\Leftrightarrow4x^2-3x+4=0\)\(\Leftrightarrow\)vô nghiệm

với \(x< -3\)\(\Rightarrow-x-3=-4x+4-1\)

\(\Leftrightarrow4x^2-4x+1-x-3=0\)

\(\Leftrightarrow4x^2-5x-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{57}}{8}\left(tm\right)\\x=\frac{5-\sqrt{57}}{8}\left(L\right)\end{cases}}\)

1.\(\frac{x+1}{x-2}=\frac{3}{4}\)

\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)

\(\Leftrightarrow4x+4=3x-6\)

<=>4x-3x=-6-4

<=>x=-10

2.\(\frac{52}{2x-1}=\frac{13}{30}\)

<=>52.30=(2x-1).13

<=>1560=26x-13

<=>-26x=-13-1560

<=>-26x=-1573

<=>x=60,5

3.\(\frac{2x-3}{x+1}=\frac{4}{7}\)

<=>(2x-3).7=(x+1).4

<=>14x-21=4x+4

<=>14x-4x=4+21

<=>10x=25

<=>x=2,5

4.\(\frac{2x+3}{42}=\frac{3x-1}{32}\)

<=>(2x+3).32=42(3x-1)

<=>64x+96=126x-42

<=>64x-126x=-42-96

<=>-62x=-138

<=>x=69/31

cho mk hỏi 

10x=25 số 10 ở đâu vậy bn 

Ta có : 3x = 2y => x/2 = y/3

7x = 5z => x/5 = z/7

 => x/2 = y/3 ; x/5 = z/7

 => x/10 = y/15 ; x/10 = z/21

 => x/10 = y/15 = z/21

 Áp dụng tính chất dãy tỉ số bằng nhau :

 x/10 = y /15 = z/21 = (x-y+z)/(10-15+21) = 32/16 = 2

đến đây xét x,y,z

 Câu b tương tự

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)

Do đó: x=20; y=30; z=42

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

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Do đó: x=18; y=16; z=15

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

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot2+3\cdot3-4}=5\)

Do đó: x-1=10; y-2=15; z-3=20

=>x=11; y=17; z=23

a,    x=10

b,   x= 2

A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)

\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)

\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)

\(\left(\frac{14}{5}x-32\right)=-60\)

\(\frac{14}{5}x=-60+32\)

\(\frac{14}{5}x=-28\)

\(x=\frac{-28}{1}:\frac{14}{5}\)

\(x=\frac{-28}{1}.\frac{5}{14}\)

\(x=\frac{-2}{1}.\frac{5}{1}=-10\)

B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)

\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)

\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)

\(2x=\frac{9}{2}-\frac{1}{2}\)

\(2x=\frac{8}{2}\)

\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)

\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)