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Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
Đặt S= \(2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+\dfrac{651-1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+1-\dfrac{1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
Đặt \(\dfrac{1}{315}=a,\dfrac{1}{651}=b\)
\(\Rightarrow S=\left(2+a\right).b-3a.\left(4-b\right)-4ab+12a\)
\(=2b+ab-12a+3ab-4ab+12a\)
\(=2b=\dfrac{2}{651}\)
a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
Đặt \(\dfrac{1}{315}=x,\dfrac{1}{651}=y\)
\(\Rightarrow A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12x\)
\(=2y+xy-12x+3xy-4xy+12x\)
\(=2y\)
Thay \(y=\dfrac{1}{651}\Rightarrow A=\dfrac{2}{651}\)
Vậy...
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\)
Đặt \(A=\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
\(=\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
\(=n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
\(=\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}=n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2}\right)=n.B\)
\(A:B=n\)
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
* là dấu nhân à bn ????
uk