\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\Leftrightarrow x^2+5x-14=x^2+3x-4\Leftrightarrow2x=10\Leftrightarrow x=5\)

x² ở đâu có vạy bạn 

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

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow x^2+15=x^2+3x-4+14\)

\(\Leftrightarrow x^2+15=x^2+3x+10\)

\(\Leftrightarrow x^2+5x-\left(x^2+3x\right)=x^2+3x+10-\left(x^2+3x\right)\)

\(\Leftrightarrow2x=10\)

=> x = 5

Giải

ta có \(\frac{x-2}{x-1}\) = \(\frac{x+4}{x+7}\)

=> (x-2)(x+7)=(x+4)(x-1)

     x²+7x-2x-14= x²-x+4x-4

     x²+5x-14-(x²-3x)=-4

(xem -4 là 1 xố hạng cần tìm của tổng)

2x-14=-4

2x=10

x=5

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

\(\Leftrightarrow\left(x-2\right)\times\left(x+7\right)=\left(x+4\right)\times\left(x-1\right)\)

\(\Leftrightarrow x^2+7x-2x-14=x^2-x+4x-4\)

\(\Leftrightarrow x^2-x^2+7x-2x+x-4x-14+4=0\)

\(\Leftrightarrow\)\(2x-10=0\)

\(\Rightarrow x=5\)

=>(x-2)(x+7)=(x+4)(x-1)=>x²+7x-2x-14=x²-x+4x-4=>3x+4x-x-x-10-4=-x+4x-4=>3x-x-10=0=>2x=10=>x=5

****

a) - 0,52 : x = - 9,36 : 16,38

- 0,52 : x = \(-\frac{4}{7}\)

\(x=\left(-0,52\right):\left(-\frac{4}{7}\right)=\frac{91}{100}=0,91\)

b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)

\(\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)

\(\frac{x}{1,61}=\frac{34}{23}\)

\(x=\frac{34}{23}.1,61=2,38\)

\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)

\(\Leftrightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)

\(\Leftrightarrow\frac{34}{23}=\frac{x}{1,61}\)

\(\Leftrightarrow x=\frac{34}{23}.1,61\)

\(\Leftrightarrow x=\frac{119}{50}\)

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 \(\frac{x}{y}=\frac{7}{9}\)

bài 2

\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+4}\)

\(\Rightarrow\) \(\left(3x+1\right)\left(5x+4\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Rightarrow15x^2+12x+10x+8=15x^2+21x-5x-7\)

\(\Rightarrow22x+8=16x-7\)

\(\Rightarrow22x+16x=-7-8\)

\(\Rightarrow6x=-15\)

\(\Rightarrow x=-2,5\)

Vậy x=-2,5

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

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

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

\(\Leftrightarrow x^2+4x-x-4=x^2+3x-2x-6\)

\(\Leftrightarrow x^2+4x-x-4-x^2-3x+2x+6=0\)

\(\Leftrightarrow2x+2=0\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)VẬY X=-1 LÀ NGHIỆM CỦA PHƯƠNG TRÌNH \(\frac{x-1}{x-2}=\frac{x+3}{x+4}\)

Ta có : \(\frac{x-1}{x-2}=\frac{x+3}{x+4}\)

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

\(\Rightarrow x^2+4x-x-4=x^2+3x-2x-6\)

\(\Rightarrow3x-4=x+6\)

\(\Rightarrow3x-x=-4+6\)

\(\Rightarrow2x=-2\)

\(\Rightarrow x=-1\)