\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

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

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

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

vậy x =....

a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

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

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn

a: =>10x-14=15-9x

=>19x=29

hay x=29/19

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x+9=32x+60

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>101x=101

hay x=1

d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)

\(\Leftrightarrow6-18x+5x-6=0\)

=>-13x=0

hay x=0

\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)

\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)

\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)

\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)

a, (0.5-\(\dfrac{2}{3}\))x=\(\dfrac{7}{12}\)

\(\Leftrightarrow\)\(\dfrac{-1}{6}\)x=\(\dfrac{7}{12}\)

\(\Leftrightarrow\)x= \(\dfrac{-7}{2}\)=-3,5

b.x \(\div\)4\(\dfrac{1}{3}\)=-2,5

\(\Leftrightarrow\) x \(\div\)\(\dfrac{13}{3}\)=-2,5

\(\Leftrightarrow\) x = -2,5 \(\times\)\(\dfrac{13}{3}\)

\(\Leftrightarrow\) x = \(\dfrac{-65}{6}\)

x.(0.5-\(\dfrac{2}{3}\))=\(\dfrac{7}{12}\)

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

x = \(\dfrac{-65}{6}\)

a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)

Bài 1:

a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)

=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)

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

=>x=2/2=1

b: \(x^2-2=0\)

=>\(x^2=2\)

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

Bài 2:

a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)

\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)\)

\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)

b: A<x<B

=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

mà \(x\in Z\)

nên \(x\in\left\{-9;-8;...;0;1\right\}\)

\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)

\(\dfrac{1}{2}.x+\dfrac{1}{2}=\dfrac{5}{2}\Leftrightarrow\dfrac{1}{2}.x=\dfrac{5}{2}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}.x=2\Leftrightarrow x=4\)

\(\dfrac{1}{2}-\dfrac{2}{3}.x=\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{1}{2}+\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{13}{12}\Leftrightarrow x=\dfrac{13}{12}:\dfrac{2}{3}\Leftrightarrow x=\dfrac{13}{8}\)

A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0

A = 0*

*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)

`3/5 : x =1/3 +1/2`

` 3/5 : x= 2/6 +3/6`

` 3/5 : x= 5/6`

` x= 3/5 : 5/6`

` x= 3/5 xx 6/5`

` x= 18/25`

__

`x: 7/15 =2`

` x= 2xx 7/15`

` x= 14/15`

__

`x-3/2=11/4-5/4`

`x-3/2= 6/4`

`x= 3/2 +3/2`

`x= 6/2`

`x=3`

__

`x+5/4 = 3/2+7/12`

`x+5/4 = 18/12+7/12`

`x+5/4 = 25/12`

`x= 25/12-5/4`

`x= 25/12- 15/12`

`x= 10/12`

`x= 5/6`

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

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