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13 tháng 9 2020

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

=> \(3x-\frac{3}{2}-3x+1=x\)

=> \(x=-\frac{1}{2}\)

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

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

=> \(\frac{4}{3}x=-\frac{9}{2}\)

=> x = \(-\frac{27}{8}\)

7/4.x+3/2=-4/5

7/4.x=-4/5-3/2

7/4.x=-23/10

x=-23/10:7/4

x=-46/35

vậy x=-46/35

1/4+3/4.x=3/4

1.x=3/4

x=3/4:1

x=3/4

vậy x=3/4

x.(1/4+1/5)-(1/7+1/8)=0

x.9/20-15/56=0

x.51/280=0

x=0:51/280

x=0

vậy x=0

3/35-(3/5+x)=2/7

(3/5+x)=3/35-2/7

(3/35+x)=-1/5

x=-1/5-3/5

x=-4/5

vậy x=-4/5

\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)

\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)

\(\frac{7}{4}.x=\frac{-7}{10}\)

\(x=\frac{-7}{10}:\frac{7}{4}\)

\(x=\frac{-2}{5}\)

\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)

\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)

\(\frac{3}{4}.x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{3}{4}\)

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

\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(x.\frac{9}{20}-\frac{15}{56}=0\)

\(x.\frac{9}{20}=\frac{15}{56}\)

\(x=\frac{15}{56}:\frac{9}{20}\)

\(x=\frac{25}{42}\)

\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

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

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

\(x=\frac{-4}{5}\)

Học tốt

3 tháng 7 2019

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

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)x+\left(x+1\right).2=\left(x+3\right)x+\left(x+3\right).5\)

\(\Leftrightarrow x^2+x+2x+2=x^2+3x+5x+15\)

\(\Leftrightarrow x^2+3x+2=x^2+8x+15\)

\(\Leftrightarrow x^2+3x-x^2-8x=15-2\)

\(\Leftrightarrow-5x=13\)

\(\Leftrightarrow x=\frac{-13}{5}\)

Vậy ...

3 tháng 7 2019

Cảm ơn bn nha 

9 tháng 8 2020

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)

\(\Leftrightarrow x=1\)

9 tháng 9 2018

a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25

b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25

c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
 

9 tháng 9 2018

Cảm ơn bạnh nha. Chúc bạn buổi tối ấm =)))) <3

phá ngoặc tính BT , nên kết quả sẽ ko ra con số nhận định !!! tui thử thui nha bà  !

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)

\(3x+y-\frac{47}{12}=\frac{1}{4}\)

\(3x+y=\frac{25}{6}\)

\(3x=\frac{25}{6}-y\)

\(x=\frac{25-25y}{18}\)

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)

\(3x+y-\frac{47}{12}=\frac{1}{4}\)

\(3x+y=\frac{25}{6}\)

\(y=\frac{25}{6}-3x\)

Vậy \(x=\frac{25-25y}{18}\)

\(y=\frac{25}{6}-3x\)

17 tháng 3 2020

Ta có:

 \(|x+\frac{1}{2}|\ge x+\frac{1}{2}\forall x;|x+\frac{1}{3}|\ge x+\frac{1}{3}\forall x;|y-5|\ge y-5\forall y;|x+\frac{1}{4}|\ge x+\frac{1}{4}\forall x\)

\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)

Mà \(|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|=\frac{1}{4}\)

\(\Rightarrow\frac{1}{4}\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)

\(\Rightarrow\frac{1}{4}\ge3x+y-\frac{47}{12}\)

\(\Rightarrow3x+y\le\frac{25}{6}\)

\(\Rightarrow x\le\frac{\frac{25}{6}-y}{3}\)

Thay vào tính y

15 tháng 7 2017

a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)

b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)

vậy x=25

15 tháng 7 2017

1.

a) \(\frac{x}{4}=\frac{16}{x^2}\)

\(\Rightarrow x^3=64\)

\(\Rightarrow x^3=4^3\)

\(\Rightarrow x=4\)

b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)

\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)

\(\frac{x}{10}=\frac{5}{2}\)

\(\Rightarrow x=\frac{5.10}{2}=25\)

2.

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}< 1\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)