Ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-\frac{432}{15}=-\frac{139}{5}\)

\(\Rightarrow\left(x+1\right).-139=5\Rightarrow x=\frac{-5}{139}-1=-\frac{144}{139}\)

Có thể bạn đã sai đề.

1 - 1/x+1 = 432/15

1/x+1 = 1- 432/15 

\(\frac{-139}{5}\)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{x\left(x+1\right)}=\frac{432}{15}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)= \(\frac{432}{15}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{432}{15}\)

\(\Leftrightarrow\frac{1}{x+1}=-\frac{139}{5}\)

\(\Leftrightarrow-139\left(x+1\right)=5\)

x=\(\frac{-144}{139}\)

Bài 4:

a: \(\dfrac{x}{15}=\dfrac{2}{3}\)

=>\(x=15\cdot\dfrac{2}{3}=10\)

b: \(x+\dfrac{-8}{13}=\dfrac{-21}{13}\)

=>\(x-\dfrac{8}{13}=-\dfrac{21}{13}\)

=>\(x=-\dfrac{21}{13}+\dfrac{8}{13}=-\dfrac{13}{13}=-1\)

c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

=>\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{1}{10}+\dfrac{5}{10}=\dfrac{6}{10}=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

d: \(\dfrac{3}{4}+\dfrac{1}{4}:x=-3\)

=>\(\dfrac{1}{4}:x=-3-\dfrac{3}{4}=-\dfrac{15}{4}\)

=>\(x=\dfrac{1}{4}:\dfrac{-15}{4}=\dfrac{-1}{15}\)

e: \(\left(\dfrac{2}{3}-x\right)\left(x-\dfrac{1}{4}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{2}{3}-x=0\\x-\dfrac{1}{4}=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{4}\end{matrix}\right.\)

f: 0,51-x=-0,35

=>x=0,51-(-0,35)

=>x=0,51+0,35=0,86

Bài 3:

a: \(-\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)

\(=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)

\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{31}{24}\)

c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)\)

\(=\dfrac{3}{5}:\dfrac{7}{20}\)

\(=\dfrac{3}{5}\cdot\dfrac{20}{7}=\dfrac{3\cdot4}{7}=\dfrac{12}{7}\)

d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)

\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}\)

\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)

e: \(\dfrac{-5}{7}+\dfrac{8}{11}+\left(-\dfrac{2}{7}\right)+\dfrac{1}{4}+\dfrac{3}{11}\)

\(=\left(-\dfrac{5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{1}{4}\)

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

f: \(\dfrac{2}{7}\cdot\dfrac{5}{13}+\dfrac{2}{7}\cdot\dfrac{6}{13}-\dfrac{2}{7}\cdot1\dfrac{11}{13}\)

\(=\dfrac{2}{7}\cdot\left(\dfrac{5}{13}+\dfrac{6}{13}-1-\dfrac{11}{13}\right)\)

\(=\dfrac{2}{7}\cdot\left(-1\right)=-\dfrac{2}{7}\)

g: \(15,3-21,5-3\cdot1,5\)

=15,3-21,5-4,5

=15,3-26

=-10,7

h: \(2\left(4^2-2\cdot4,1\right)+1.25:5\)

=2(16-8,2)+0,25

=15,6+0,25

=15,85

/x-3/=0

x-3=0

x=3

Where did you go on holiday ?

where did you go on holiday?

_Where did you go on holiday?

\(\dfrac{15}{19}\times\dfrac{25}{33}-\dfrac{15}{19}\times\dfrac{14}{33}\)

\(=\dfrac{15}{19}\times\left(\dfrac{25}{33}-\dfrac{14}{33}\right)\)

\(=\dfrac{15}{19}\times\dfrac{11}{33}\)

\(=\dfrac{5}{19}\)

Lần sau để đúng môn học nha bạn.

Thế có biết câu hỏi này ko

2300x10=?

3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25 

= 3/25 x 13/7 + 3/7 x 1/25 

= (3 x 13/7 + 3/7 ) x 1/25 

= 42/7 x 1/25 

= 6 x 1/25 

= 6/25

\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)

\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)

\(=\dfrac{39}{175}+\dfrac{3}{175}\)

\(=\dfrac{39+3}{175}\)

\(=\dfrac{42}{175}\)

\(=\dfrac{6}{25}\)

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{15}{7}\) + \(\dfrac{3}{25}\) \(\times\) \(\dfrac{1}{7}\) - \(\dfrac{2}{7}\times\) \(\dfrac{3}{25}\)

A = \(\dfrac{3}{25}\) \(\times\)( \(\dfrac{15}{7}\) + \(\dfrac{1}{7}-\dfrac{2}{7}\))

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{14}{7}\)

A = \(\dfrac{6}{25}\)

f(100)=100⁸-101.100⁷+101.100⁶-101.100⁵+...+101.100²-101.100+5

=100⁸-(100+1)100⁷+(100+1)100⁶-(100+1)100⁵+...+(100+1)100²-(100+1)100+25

=100⁸-100⁸-100⁷+100⁷+100⁶-100⁶-100⁵+...+100³+100²-100²-100+25

=0-100+25=-75