K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1 tháng 7 2019

a) \(\left(-\frac{1}{5}\right)^2:x=\frac{1}{2,5}\)

=> \(\frac{1}{25}:x=\frac{2}{5}\)

=> \(x=\frac{1}{25}:\frac{2}{5}\)

=> \(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\).

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!

2 tháng 7 2019

Cũng cảm mơn bn nhé! Đã giúp mih 😁😊😆

1 tháng 7 2019
https://i.imgur.com/gSf9gNJ.jpg

chữ đẹp khó tin

a) 9x-1/4=3/2

=>9x=3/2+1/4

=>9x=7/4

=>x=7/4:9

=>x=7/36

Vậy x=7/36

b)(4x+2):2,5=3,2:0,5

=>(4x+2):2,5=6,4

=>4x+2=6,4.2,5

=>4x+2=16

=>4x=16-2

=>4x=14

=>x=14:4

=>x=7/2

Vậy x=7/2

c) 5,4/x-2=6/7

=>5,4/x=6/7+2

=>5,4/x=20/7

=>x=5,4 :20/7

=>x=1,89

Vậy x= 1,89

d) 0,5:2=3:(2x+7)

=>3:(2x+7)=0,25

=>2x+7=3:0,25

=>2x+7=12

=>2x=12-7

=>2x=5

=>x=5/2

Vậy x=5/2

20 tháng 7 2023

a) 9x-1/4=3/2

=>9x=3/2+1/4

=>9x=7/4

=>x=7/4:9

=>x=7/36

Vậy x=7/36

b)(4x+2):2,5=3,2:0,5

=>(4x+2):2,5=6,4

=>4x+2=6,4.2,5

=>4x+2=16

=>4x=16-2

=>4x=14

=>x=14:4

=>x=7/2

Vậy x=7/2

c) 5,4/x-2=6/7

=>5,4/x=6/7+2

=>5,4/x=20/7

=>x=5,4 :20/7

=>x=1,89

Vậy x= 1,89

d) 0,5:2=3:(2x+7)

=>3:(2x+7)=0,25

=>2x+7=3:0,25

=>2x+7=12

=>2x=12-7

=>2x=5

=>x=5/2

Vậy x=5/2

28 tháng 6 2019

Câu 1 :

\(a,2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)

\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)

\(\Rightarrow7x=\frac{3}{2}-\frac{4}{5}\)

\(\Rightarrow7x=\frac{7}{10}\)\(\Leftrightarrow x=0,1\)

\(b,\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)

\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)

\(\Rightarrow11x=\frac{2}{3}+1-\frac{3}{2}\)

\(\Rightarrow11x=\frac{4+6-9}{6}-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{66}\)

28 tháng 6 2019

Câu 2 :

\(a,\frac{2}{x-1}< 0\)

Vì \(2>0\Rightarrow\)để \(\frac{2}{x-1}< 0\)thì \(x-1< 0\Leftrightarrow x< 1\)

\(b,\frac{-5}{x-1}< 0\)

Vì \(-5< 0\)\(\Rightarrow\)để \(\frac{-5}{x-1}< 0\)thì \(x-1>0\Rightarrow x>1\)

\(c,\frac{7}{x-6}>0\)

Vì \(7>0\Rightarrow\)để \(\frac{7}{x-6}>0\)thì \(x-6>0\Rightarrow x>6\)

28 tháng 3 2020

\(a,x.\frac{-3}{7}=\frac{4}{21}\)

\(x=\frac{4}{21}:\frac{-3}{7}\)

\(x=\frac{-4}{9}\)

\(b,\frac{-4}{7}:x=\frac{2}{5}\)

\(x=\frac{-4}{7}:\frac{2}{5}\)

\(x=\frac{-10}{7}\)

\(c,x+\frac{1}{12}=\frac{-3}{8}\)

\(x=\frac{-3}{8}-\frac{1}{12}\)

\(x=\frac{-11}{24}\)

\(d,\frac{2}{15}-x=\frac{-3}{10}\)

\(x=\frac{2}{15}+\frac{3}{10}\)

\(x=\frac{13}{30}\)

28 tháng 3 2020

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

\(-x=\frac{-3}{10}\)

\(x=\frac{3}{10}\)

\(f,\frac{3}{4}.\left(x+1\right)-\frac{1}{2}=\frac{3}{7}\)

\(\frac{3}{4}.\left(x+1\right)=\frac{13}{14}\)

\(x+1=\frac{26}{21}\)

\(x=\frac{5}{21}\)

\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)

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

\(2x=\frac{-3}{2}+\frac{11}{4}\)

\(2x=\frac{-17}{4}\)

\(x=\frac{-17}{8}\)

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

\(-x=\frac{-3}{10}\)

\(x=\frac{3}{10}\)

chúc bạn học tốt !!!

17 tháng 10 2020

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

=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)

=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)

=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)

Vậy x = 2/5

b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)

=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)

=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)

Vậy x = -6/11

c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

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

=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)

=> \(\frac{2}{3}-\frac{4}{3}x=5\)

=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)

Vậy x = -13/4

d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)

=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)

=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)

=> \(3\left(229+45x\right)=4\left(61+30x\right)\)

=> \(687+135x=244+120x\)

=> \(687+135x-244-120x=0\)

=> \(\left(687-244\right)+\left(135x-120x\right)=0\)

=> \(443+15x=0\)

=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)

Vậy x = -443/15

15 tháng 6 2016

1/ a/\(-\frac{7}{18}=\left(-\frac{7}{2}\right)\left(\frac{1}{9}\right)\)

b/\(-\frac{7}{18}=\left(-\frac{7}{9}\right):2\)

2/

a/\(\frac{7}{15}\cdot\left(-\frac{3}{8}-\frac{3}{7}\right)=\frac{7}{15}\cdot\left(-\frac{45}{56}\right)=-\frac{3}{8}\)

b/\(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+-\frac{4}{4}\right):\frac{3}{7}\)

\(=\left(-\frac{7}{20}\right):\frac{3}{7}+\left(-\frac{2}{5}\right):\frac{3}{7}\)

\(=\left(-\frac{49}{60}\right)+\left(-\frac{14}{15}\right)=-\frac{7}{4}\)

c/\(\frac{2}{3}\cdot\left(-\frac{5}{2}\right)+\frac{10}{15}\cdot\left(-\frac{3}{7}\right)-\frac{2}{3}\cdot\left(-\frac{5}{3}\right)\)

\(=\frac{2}{3}\cdot\left(-\frac{5}{2}-\frac{3}{7}+\frac{5}{3}\right)=-\frac{53}{63}\)

3/

\(2-\left(3-x\right)=-\frac{3}{2}\)

\(2-3+x=-\frac{3}{2}\)

\(x=-\frac{3}{2}+3-2=-\frac{1}{2}\)

4/ 

a/ Ta có 2 trường hợp:

TH1: \(x-3,5=7,5\)

\(x=7,5+3,5=11\)

TH2: \(x-3,5=-7,5\)

\(x=-7,5+3,5=-4\)

b/ Ta có 2 trường hợp:

TH1:\(x-0,4=3,6\)

\(x=4\)

TH2: \(x-0,4=-3,6\)

\(x=-3.2\)

c/ Ta có 2 trường hợp:

TH1:\(x+\frac{4}{5}=\frac{3}{2}\)

\(x=\frac{7}{10}\)

TH2:\(x+\frac{4}{5}=-\frac{3}{2}\)

\(x=-\frac{32}{10}\)

15 tháng 6 2016

cam on  

12 tháng 9 2021

\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)