\(\frac{x-1}{21}=\frac{3}{x+1}\)

=> \(\left(x-1\right)\left(x+1\right)=21\cdot3\)

=> \(x^2-1=63\)

=> \(x^2=64\)

=> \(\orbr{\begin{cases}x^2=8^2\\x^2=\left(-8\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)

\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

=> \(\frac{12}{13}x=2\)

=> \(x=\frac{13}{6}\)

d, \(\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow x^2-1=63\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

e, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)

 mik ko chép lại đề, mik làm luôn: 

a)  x - \(\frac{31}{36}=\frac{-13}{38}\)

x = \(\frac{-13}{18}+\frac{31}{36}\)

\(x=\frac{5}{36}\)

b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)

\(\frac{11}{7}-x=\frac{3}{7}\)

x = \(\frac{11}{7}-\frac{3}{7}\)

x = 8/7

c) x + 3/11 = 23/44

x = 23/44 - 3/11

x = 1/4

d) \(\frac{1}{12}-x=\frac{-11}{9}\)

x = \(\frac{1}{12}+\frac{11}{9}\)

x = 47/36

e) \(x-\frac{2}{3}=\frac{-17}{3}\)

x= -17/3 + 2/3

x = -5 

f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)

x - 1/2 = 3/4

x = 3/4 + 1/2 

x = 5/4

g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)

2x + 3/8 = -3 / 8

2x = -3/8 - 3/8 

2x = -9/8

x = -9/8.1/2 

x = -9/16

h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)

x  - \(\frac{x}{3}=\frac{1}{12}\)

x = \(\frac{1}{12}+\frac{x}{3}\)

x = \(\frac{1+4x}{12}\)

=> 12x = 1+4x

12x - 4x = 1

8x = 1

x = 1/8 

Trả lời nhanh gọn lẹ nhé, mình k cho :)

a) \(x-\frac{5}{12}-\frac{4}{9}=\frac{-13}{18}\)

  \(x-\frac{5}{12}=\frac{-13}{18}+\frac{4}{9}\)

  \(x-\frac{5}{12}=\frac{-13}{18}+\frac{8}{18}\)

  \(x-\frac{5}{12}=\frac{-5}{18}\)

 \(x=\frac{-5}{18}+\frac{5}{12}\)

....

\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)

\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)

\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)

\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{3}\)

\(\Rightarrow x=3\)

Vậy \(x=3\)

\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)

\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)

\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)

\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)

\(\Rightarrow x=19\)

Vậy \(x=19\)

#Mạt Mạt#

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CẢM ƠN VƯƠNG MẠT MẠT NHA!

\(a.\frac{4}{3}-\frac{3}{2}:X=\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{4}{3}-\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{8}{6}-\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{7}{6}\)

               \(X=\frac{3}{2}:\frac{7}{6}\)

              \(X=\frac{3}{2}\times\frac{6}{7}\)

              \(X=\frac{9}{7}\)

\(b.\left(X+\frac{2}{3}\right):\frac{1}{3}=\frac{41}{3}\)

    \(X-\frac{2}{3}=\frac{41}{3}.\frac{1}{3}\)

   \(X-\frac{2}{3}=\frac{41}{9}\)

  \(X=\frac{41}{9}+\frac{2}{3}\)

  \(X=\frac{41}{9}+\frac{6}{9}\)

 \(X=\frac{47}{9}\)

\(c.X\times\frac{11}{5}-X\times\frac{4}{7}=\frac{5}{7}\)

  \(X\times\left(\frac{11}{5}-\frac{4}{7}\right)=\frac{5}{7}\)

  \(X\times\frac{57}{35}=\frac{5}{7}\)

  \(X=\frac{5}{7}:\frac{57}{35}\)

 \(X=\frac{5}{7}\times\frac{35}{57}\)

\(X=\frac{25}{57}\)

a) Dễ thấy VT > 0;mà VT=VP

=>VP > 0 => 4x > 0=> x > 0

=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)

=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)

\(=>3x+1=4x=>x=1\)

a)  Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )

Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)

=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)

<=>x=1

Vậy x=1

b)Điều kiện: \(x\ne-3;-10;-21;-34\)

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

=>x+34-x-3=x

<=>x=31 (nhận)

Vậy x=31

a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)

Ta có: \(\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{3}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\end{cases}\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|\ge0\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}\)

Khi đó, ta có: \(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)

\(\Rightarrow3x+1=4x\)

\(\Rightarrow x=1\)

b) Từ đề suy ra:

\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{x+34}{\left(x+3\right)\left(x+34\right)}-\frac{x+3}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow x=31\)

a Đ

b S

c S

d Đ

a ) S 

b ) Đ

c ) S

d ) Đ

k cho mk nhé 

Mai Hồng Ngọc? Vũ Thị Thu Hằng? Ai đúng dzậy -_-*

\(\left(x-\frac{7}{3}\right):2\frac{3}{21}+\frac{3}{5}=0,16\)

<=> \(\left(x-\frac{7}{3}\right):\frac{45}{21}+\frac{3}{5}=\frac{4}{25}\)

<=> \(\left(x-\frac{7}{3}\right):\frac{15}{7}=-\frac{11}{25}\)

<=> \(x-\frac{7}{3}=\frac{-33}{35}\)

<=> \(x=\frac{146}{105}\)

\(\left(x+\frac{5}{6}\right).2\frac{2}{5}-1\frac{1}{4}=0,35\)

<=> \(\left(x+\frac{5}{6}\right).\frac{12}{5}-\frac{5}{4}=\frac{7}{20}\)

<=> \(\left(x+\frac{5}{6}\right).\frac{12}{7}=\frac{8}{5}\)

<=> \(x+\frac{5}{6}=\frac{14}{15}\)

<=> \(x=\frac{1}{10}\)

học tốt

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)

\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)

\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)

\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)

\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)

=> 5x - 20 = 2

=> 5x = 22 

\(\Rightarrow x=\frac{22}{5}=4,4\)

Vậy, x = 4,4