1. <=>15 (2x+4)= 7(4x-2)

<=>30x+60= 28x-14

<=>2x= -74

<=>x= -37

2. <=> -5(12-7x)= -13(4-3x)

<=> -60+35x = -52+39x

<=> -4x= 8

<=> x= -2

3.<=> 88x-16 = 77x+55

<=> 11x= 71

<=> x=71/11

\(\frac{2x}{3}=\frac{7x-3}{15}\)

\(\Rightarrow30x=3\left(7x-3\right)\)

\(30x=21x-9\)

\(9x=-9\)

\(x=-1\)

\(\frac{11x-2}{7x+5}=\frac{11}{8}\)

\(\Rightarrow8\left(11x-2\right)=11\left(7x+5\right)\)

\(88x-16=77x+55\)

\(11x=71\)

\(x=\frac{71}{11}\)

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

\(\Rightarrow4\left(2x+3\right)=3\left(3x+1\right)\)

\(8x+12=9x+3\)

\(-x=-9\)

\(x=9\)

mk lộn đề câu a

bn giải lại giùm mk nha

\(\frac{2x+3}{6}=\frac{7x-3}{15}\)

\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)

\(\Rightarrow x^2+8x+12=32\)

\(\Leftrightarrow x^2+8x-20=0\)

Đến đây đơn giản rồi nhé

\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=0\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=0\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{x+6}\)

\(\Leftrightarrow x+6=x+2\)

\(\Leftrightarrow x-x=2-6\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm

@ Harley @ ơi sai rồi, phải quy đồng như bạn Vân Nhi chứ không thể rút tử như thế

a)

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

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

\(\Leftrightarrow7x-21=25+5x\)

\(\Leftrightarrow7x-5x=25+21\)

\(\Leftrightarrow2x=46\)

\(\Leftrightarrow x=23\)

b) 

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

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=7.9\)

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

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow x=8\)

Mẫy bài còn lại làm tương tự 

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

\(\Leftrightarrow(x-1)(x+3)=(x-2)(x+2)\)

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

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

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

\(\Leftrightarrow2x-3=-4\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

\(d,\frac{11x-2}{7x+5}=\frac{11}{8}\)

\(\Leftrightarrow8(11x-2)=11(7x+5)\)

\(\Leftrightarrow88x-16=77x+55\)

\(\Leftrightarrow88x-16-77x=55\)

\(\Leftrightarrow88x-77x-16=55\)

\(\Leftrightarrow11x=71\Leftrightarrow x=\frac{71}{11}\)

ĐK: \(x\ne-2;-3;-4;-5;-6\)

\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\Leftrightarrow\left(x+2\right)\left(x+6\right)=32\)

\(\Leftrightarrow x^2+8x-20=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

\(...\Leftrightarrow\frac{1}{\left(x+2\right) \left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{18}\Leftrightarrow\frac{x+6}{\left(x+2\right)\left(x+6\right)}-\frac{x+2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\Rightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=72\)

=> \(x^2+8x-60=0\)

Phân tich đa thức thành nhân tử để tìm x

pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8

<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8

<=> 1/x+2 - 1/x+6 = 1/8

<=> (x+6-x-2)/(x+2).(x+6) = 1/8

<=> 4/(x+2).(x+6) = 1/8

<=>(x+2).(x+6) = 4 : 1/8 = 32

<=>x^2 + 8x + 12 = 32

<=> x^2+8x+12-32=0

<=>x^2+8x-20=0

<=>(x-2).(x+10)=0

<=> x-2 =0 hoặc x+10 = 0

<=> x=2 hoặc x=-10

giang sinh an lanh $%###Xuyen gam cu chuoi###%$

1) \(\frac{2x+3}{6}=\frac{7x-3}{-5}\)

  \(-5\left(2x+3\right)=6\left(7x-3\right)\)

 \(-10x-15=42x-18\)

 \(52x=3\)

 \(x=\frac{3}{52}\)

2) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)

\(-5\left(12-7x\right)=-13\left(4-3x\right)\)

\(-60+35x=-52+39x\)

\(39x-35x=-60+52\)

\(4x=-8\)

\(x=-2\)