\(\frac{4}{5}+\frac{1}{6}-\frac{4}{9}=\frac{29}{30}-\frac{4}{9}=\frac{87}{90}-\frac{40}{90}=\frac{11}{30}\)

\(\frac{2}{3}-\frac{1}{4}=\frac{8}{12}-\frac{3}{12}=\frac{5}{12}\)

\(\frac{5}{12}=\frac{150}{360};\frac{11}{30}=\frac{132}{360}\)

\(x=\frac{19}{360}\)

a) 3/8 . x = 9/8 - 1

3/8 . x = 1/8

x = 1/8 : 3/8

x = 1/3

b) 4/5 . x = 7/5 - 1/5

4/5 . x = 6/5

x = 6/5 : 4/5

x = 3/2

c) 12/7 : x + 2/3 = 7/5

12/7 : x = 7/5 - 2/3

12/7 : x = 11/15

x = 12/7 : 11/15

x = 180/77

d) 3.(x + 7) - 15 = 27

3.(x + 7) = 27 + 15

3.(x + 7) = 42

x + 7 = 42 : 3

x + 7 = 14

x = 14 - 7

x = 7

a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)

b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)

\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)

c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)

d) \(3\left(x+7\right)-15=27\)

\(\Leftrightarrow3\left(x+7\right)=42\)

\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)

bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;

mik kiểm tra rùi

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)

hong ai giúp mik dọ? ;-;

a)\(-\frac{12}{7}\cdot\left(\frac{3}{4}-x\right)\cdot\frac{1}{4}=0\)

=>\(\frac{3}{4}-x=0\)

=>\(x=\frac{3}{4}\)

 \(x=\frac{3}{4}\)

b) \(x:\frac{17}{8}=-\frac{2}{5}\cdot\left(-\frac{9}{17}\right)\)

=>\(x:\frac{17}{8}=\frac{18}{85}\)

=>\(x=\frac{18}{85}\cdot\frac{17}{8}=\frac{9}{20}\)

 \(x=\frac{9}{20}\)

a.7,2 :2 x 58,6 + 2,93 x 2 x 64

= 3,6 x 58,6 + 5,86 x 64

=210,96 + 357,04

=586

b.\(4\frac{2}{5}+2\frac{3}{7}-2\frac{2}{5}+5\frac{4}{7}\)

=\(\left(4\frac{2}{5}-2\frac{2}{5}\right)+\left(2\frac{3}{7}+5\frac{4}{7}\right)\)

= 2 + 8

= 10

c. 2 x 41 x 36 + 8 x 9 x 58 + 12 x 6

=72 x 41 + 72 x 58 + 72 x 1

=72 x ( 41 + 58 +1 )

=72 x 100

=7200

d, \(\frac{3}{5}\)x \(\frac{1}{2}\):\(\frac{2}{5}\)

=\(\frac{3}{5}\)x \(\frac{1}{2}\)x\(\frac{5}{2}\)

=\(\frac{3}{4}\)

phá ngoặc ra mà tính đi bn ơi

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

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

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)

b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)

\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)

c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)

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

\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)

\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)

d. \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7=x\)

Vậy x = 7

e. \(\left(x-1\right)-\left(2x-1\right)=9\)

\(\Leftrightarrow x-1-2x+1=9\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

g. \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

a) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)

=> \(\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

=> \(x=2:\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

b) \(x:\frac{13}{3}=-2,5\)

=> \(x:\frac{13}{3}=-\frac{5}{2}\)

=> \(x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

c) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

=> \(\frac{4x-3}{12}=-\frac{10}{12}\)

=> 4x - 3 = -10

=> 4x = -10 + 3 = -7

=> x = -7/4

Bài 2 :

\(A=a\cdot\frac{1}{3}+a\cdot\frac{1}{4}-a\cdot\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\cdot\frac{5}{12}\)

Thay a = -3/5 vào biểu thức ta có : \(A=\left(-\frac{3}{5}\right)\cdot\frac{5}{12}=\frac{-3}{12}=\frac{-1}{4}\)

\(B=b\cdot\frac{5}{6}+b\cdot\frac{3}{4}-b\cdot\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\cdot\frac{13}{12}\)

Thay b = 12/13 vào ta được kết quả là 1

a ) \(\frac{25}{9}-\frac{12}{13}\cdot x=\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}\cdot x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

\(\Rightarrow x=2\div\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

Vậy ...

b ) \(x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

Vậy ..

c ) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

\(\Rightarrow\frac{4x-3}{12}=-\frac{10}{12}\)

\(\Rightarrow4x-3=-10\)

\(\Rightarrow4x=-10+3=-7\)

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

Vậy ....

Bài 2 : 

\(A=a\cdot\frac{1}{3}+a\cdot\frac{1}{4}-a\cdot\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}=a\cdot\frac{5}{12}\right)\)

Thay \(a=-\frac{3}{5}\)vào biểu thức ta được : \(A=\left(-\frac{3}{5}\right)\cdot\frac{5}{12}=-\frac{3}{12}=-\frac{1}{4}\)

\(B=b\cdot\frac{5}{6}+b\cdot\frac{3}{4}-b\cdot\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\cdot\frac{13}{12}\)

Thay \(b=\frac{12}{13}\)vào biểu thức , dễ thấy kết quả bằng 1 .