Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

mình đang cần gấp câu trả lời ,bạn nào giải được nhanh k luôn ,hứa đấy

bạn đi học thêm à ?

1) -5+(-3+12-x)=7-(2-8)

=>-5+(9-x)=7-(-6)

=>-5+9-x=13

=>x=9-5-13=-9

2) 8-(7-x+3)=-|-7+2|-(-9)

=>8-(10-x)=-|-5|+9

=>8-10+x=-5+9

=>x=-5+9-8+10=6

3)-(12-5+9-x)=3-|5-8+2|

=>-(16-x)=3-|-1|

=>-16+x=3-1

=>x=3-1+16=18

4)-(-3-x+8)-(-3+5)=|5-8|

=>-(5-x)-2=|-3|

=>-5+x-2=3

=>x=3+5+2=10

5)-|5-8|-(4-x+5)=4+|3-9|

=>-|-3|-(9-x)=4+|-6|

=>-3-9+x=4+6

=>x=4+6+3+9=22

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

1; |\(x\) + 2| = 4

    \(\left[{}\begin{matrix}x+2=-4\\x+2=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-4-2\\x=4-2\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 6; 2}

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

Ko chép đề

\(2)x-17+x=x-7\)

\(x+x-x=-7+17\)

\(x=-10\)

Vậy ....

âm 10 , nhỉ ?

a, \(\frac{2}{3}:\frac{5}{7}:\frac{x}{9}=\frac{2}{7}:\frac{3}{5}:\frac{10}{9}\Leftrightarrow\frac{2\times7\times9}{3\times5\times x}=\frac{2\times5\times9}{7\times3\times10}\Leftrightarrow\frac{42}{5\times x}=\frac{3}{7}\Leftrightarrow294=15\times x\Leftrightarrow x=19,6\)

Các phần sau tương tự.

Dùng phương pháp nhân chéo và kiểm tra trình độ tính toán, khả năng rút gọn.

Thay kết quả vào x để kiểm tra lại bài.

cho hỏi Xx9/7:3/4=12/15

a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)

\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)

\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)

\(\dfrac{x}{3}=\dfrac{29}{15}\)

\(x=5,8\)

b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)

\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)

\(2x-1=\dfrac{5}{2}\)

\(2x=\dfrac{7}{2}\)

\(x=\dfrac{7}{4}\)

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad