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17 tháng 3

a) \(x:2+x:3=x-35\)

\(x-x:2-x:3=35\)
\(x\times1-x\times\dfrac{1}{2}-x\times\dfrac{1}{3}=35\)

\(x\times\left(1-\dfrac{1}{2}-\dfrac{1}{3}\right)=35\)

\(x\times\dfrac{1}{6}=35\)

\(x=35:\dfrac{1}{6}\)

\(x=210\)

b) \(\left(\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}\right)\times x=\dfrac{1}{3}\)

\(\left(\dfrac{2}{2\times4}+\dfrac{2}{4\times6}+\dfrac{2}{6\times8}+\dfrac{2}{8\times10}\right)\times x=\dfrac{2}{3}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\times x=\dfrac{2}{3}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\times x=\dfrac{2}{3}\)

\(\dfrac{3}{10}\times x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:\dfrac{3}{10}\)

\(x=\dfrac{20}{9}\)

31 tháng 12 2022

a: =>x*2/15=2/7

=>x=2/7:2/15=2/7*15/2=15/7

b: x=3:7/5=15/7

c: x=-1/2:4/9=-1/2*9/4=-9/8

d: x=-8/3:3/8=-64/9

g: =>4/11x=2/5+1/3=6/15+5/15=11/15

=>x=11/15:4/11=121/60

l: =>1/4:x=1-3/2=-1/2

=>x=-1/4:1/2=-1/4*2=-1/2

k: =>x:7=-1/3+5/2=-2/6+15/6=13/6

=>x=91/6

21 tháng 2 2023

a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)

b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)

c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)

21 tháng 2 2023

giúp mik với ạ, mình sẽ tick ạ! Thanks.

3 tháng 5 2023

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

3 tháng 5 2023

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)\(\times\) \(\dfrac{8}{7}\): 2

\(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

\(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

\(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)\(\times\) ( 1 - \(\dfrac{1}{3}\)\(\times\) ( 1 - \(\dfrac{1}{4}\)\(\times\) ( 1 - \(\dfrac{1}{5}\))

\(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

\(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

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

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

1/* Chứng minh rằng:

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+..+\dfrac{1}{50}\)

2/* Cho:

A=\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+.....+\dfrac{1}{99\times100}\). Chứng minh rằng:\(\dfrac{7}{12}< A>\dfrac{5}{6}\)

Các bn giúp mk những bài này nha!

4
16 tháng 7 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

16 tháng 7 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

25 tháng 9 2023

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

8 tháng 10 2021

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

\(\left(x+4\right)\left(x+4\right)=100\)

\(\left(x+4\right)^2=10^2\)

\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

8 tháng 10 2021

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

1 tháng 1 2018

a) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\left(\dfrac{-8}{5}+x\right).\dfrac{13}{12}=\dfrac{13}{6}\)

\(-\dfrac{8}{5}+x=\dfrac{13}{6}:\dfrac{13}{12}\)

\(-\dfrac{8}{5}+x=2\)

\(x=2+\dfrac{8}{5}\)

\(x=\dfrac{18}{5}\)

b) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}\)

\(-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)\)

\(x=\dfrac{413}{160}\)

1 tháng 1 2018

a, \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

=> \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)

=> \(\left(-1\dfrac{3}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)

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

=> \(\dfrac{-8}{5}+x=2\)

=> x= \(2+\dfrac{8}{5}=\dfrac{10}{5}+\dfrac{8}{5}\)

=> x= \(\dfrac{18}{5}\)

AH
Akai Haruma
Giáo viên
13 tháng 12 2021

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

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

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

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

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

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

 

26 tháng 7 2017

a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)

\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)

\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)

\(=\left(-\dfrac{1}{2}\right)2+1\)

\(=-1+1\)

\(=0\)

@Trịnh Thị Thảo Nhi

29 tháng 4 2018

a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1

=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1

=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1

=(−12)2+1=(−12)2+1

=−1+1=−1+1

=0=0