Chọn B

Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2

1 , 10

2 , 8

3 , 5

4 , 2

5 , 1

6 ,

1/ `|x|=10<=> x=\pm 10`

2/ `|x-8|=0<=>x-8=0<=>x=8`

3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`

4/ `|x+1|=3`

$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$

5/ `15-x=16-(14-42)`

`<=>15-x=16+28`

`<=>15-x=44`

`<=>x=-29`

6/ `210-(x-12)=168`

`<=>210-x+12=168`

`<=>222-x=168`

`<=>x=54`

1.

\(\left|x\right|=10\Leftrightarrow x=\pm10\)

2.

\(\left|x-8\right|=0\Leftrightarrow x-8=0\Leftrightarrow x=8\)

3.

\(7+\left|x\right|=12\Leftrightarrow\left|x\right|=5\Leftrightarrow x=\pm5\)

\(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow24x+10=0\)

\(\Leftrightarrow x=-\dfrac{5}{12}\)

chỉ

có

1

từ

để

miêu

tả

dễ

\(\frac{3}{\frac{1}{4}}x-\frac{1}{\frac{2}{3}}=\frac{5}{12}-\frac{7}{6}x\)

\(3:\frac{1}{4}x-1:\frac{2}{3}=\frac{5}{12}-\frac{7}{6}x\)

\(1x-1.5=\frac{5}{12}-\frac{7}{6}x\)

\(1x+\frac{7}{6}x=1.5+\frac{5}{12}\)

\(\left(1+\frac{7}{6}\right)x=\frac{18}{12}+\frac{5}{12}=\frac{23}{12}\)

\(\frac{13}{6}x=\frac{23}{12}\)

\(x=\frac{23}{12}:\frac{13}{6}\)

\(x=\frac{23}{12}\cdot\frac{6}{13}=\frac{23}{26}\)

ai ko biết thì đừng milk

1.Tìm x , biết

.2x -1/2-1/6-1/12-...- 1/49*50=7-1/50+x

=> 2x- ( 1/2+1/6+1/12+...1/ 49.50 )= 7-1/50+x

=> 2x -( 1/1.2 + 1/2.3+1/3.4+...+1/49.50)= 7-1/50+x

=> 2x - ( 1- 1/2+ 1/2-1/3+1/3-1/4+...+1/49-1/50) = 7-1/50 + x

=> 2x - ( 1-1/50) =7-1/50 + x

=> 2x- 1+ 1/50=7-1/50+ x

=> 1+1/50= 2x- (7 - 1/50+ x)

=> 1+1/50 = 2x- 7 + 1/50- x

=> 1+1/50 = x + 1/50 - 7

=> 1 = x + 1/50 - 7 - 1/50

=> 1 = x - 7

=> x = 8 

Vậy...

Tham khảo thêm:Câu hỏi của Cừu beta - Toán lớp 7 - Học toán với OnlineMath

LinkCâu hỏi của Cừu beta - Toán lớp 7 - Học toán với OnlineMath

\(\text{2x -1/2-1/6-1/12-...- 1/49*50=7-1/50+x }\)

=> \(\text{2x- ( 1/2+1/6+1/12+...1/ 49.50 )= 7-1/50+x }\)

=> \(\text{2x -( 1/1.2 + 1/2.3+1/3.4+...+1/49.50)= 7-1/50+x }\)

=>\(\text{ 2x - ( 1- 1/2+ 1/2-1/3+1/3-1/4+...+1/49-1/50) = 7-1/50 + x}\)

=> \(\text{2x - ( 1-1/50) =7-1/50 + x }\)

=>\(\text{ 2x- 1+ 1/50=7-1/50+ x }\)

=> \(\text{1+1/50= 2x- (7 - 1/50+ x) }\)

=>\(\text{ 1+1/50 = 2x- 7 + 1/50- x}\)

=>\(\text{ 1+1/50 = x + 1/50 - 7 }\)

=>\(\text{ 1 = x + 1/50 - 7 - 1/50 }\)

=> \(\text{1 = x - 7}\)

=> \(\text{x = 8 }\)

\(a,\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}:\dfrac{11}{12}\)

\(\Leftrightarrow x=-\dfrac{11}{12}.\dfrac{12}{11}\)

\(\Leftrightarrow x=-1\)

\(b,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{2}{3}.\left(\dfrac{1}{6}-x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow3-\dfrac{1}{9}+\dfrac{2}{3}x=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}-3+\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{20}{9}\)

\(\Leftrightarrow x=-\dfrac{20}{9}:\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{10}{3}\)

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

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

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

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

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

\(x=\dfrac{-1}{4}\)

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=9\)

\(\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\)

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

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).