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1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2016}\left(1+2+...+2016\right)\)\(=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{2016.2017}{2016.2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2017}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2017}{2}\)
\(=\dfrac{1}{2}\left(2+3+...+2017\right)\)
Đặt \(A=2+3+...+2017\)
\(=2017+2016+...+2\)
\(\Rightarrow2A=\left(2+2017\right)+\left(3+2016\right)+...+\left(2017+2\right)\) ( 2016 cặp số )
\(\Rightarrow2A=2019+2019+...+2019\) ( 2016 số )
\(\Rightarrow2A=4070304\)
\(\Rightarrow A=2035152\)
\(\Rightarrow P=1017576\)
Vậy...
P= 1+1/2.3+1/3.6+...+1/2016.2033136
P= 1+3/2+2+...+2017/2
P= 2/2+3/2+4/2+...+2017/2
P=\(\dfrac{2+3+4+...+2017}{2}\)
P= \(\dfrac{2035152}{2}\)
P= 1017576
a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)
= \(\dfrac{-7}{9}.\dfrac{11}{4}\)
= \(\dfrac{-77}{36}\)
b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)
= \(\dfrac{2}{3}+\dfrac{-2}{15}\)
= \(\dfrac{10}{15}+\dfrac{-2}{15}\)
= \(\dfrac{-8}{15}\)
c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)
= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)
= \(\dfrac{2}{3}-5\)
= \(\dfrac{-13}{3}\)
d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)
= \(\dfrac{1}{2}\) . 11 - 7
= \(\dfrac{11}{2}-\dfrac{14}{2}\)
= \(\dfrac{-3}{2}\)
e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)
= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)
= \(\dfrac{3}{4}.-28\)
= \(-21\)
các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho
a: \(A=\left(\dfrac{-3}{4}+\dfrac{-2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{3}{5}\right)+\dfrac{1}{57}\)
\(=\dfrac{-27-8-1}{36}+\dfrac{5+1+9}{15}+\dfrac{1}{57}\)
=1/57
b: \(B=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+1+2}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
=1/41
c: \(C=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{107}\)
=1-1+1/107
=1/107
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...................+\dfrac{1}{3^{2016}}\)
\(3A=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...............+\dfrac{1}{3^{2016}}\right)\)
\(3A=\dfrac{3}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+.................+\dfrac{3}{3^{2016}}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+......................+\dfrac{1}{3^{2015}}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{2015}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+.............+\dfrac{1}{3^{2016}}\right)\)
\(A=1-\dfrac{1}{3^{2016}}\)
\(A=\dfrac{3^{2016}-1}{3^{2016}}\)
~~ Chúc bn học tốt ~~
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\)
\(\dfrac{1}{3}A=\dfrac{1}{3}.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\right)\)
\(\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2017}}\)
\(\dfrac{1}{3}A-A=\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2017}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}\right)\)
\(-\dfrac{2}{3}A=\dfrac{1}{3^{2017}}-\dfrac{1}{3}\)
\(A=\dfrac{\dfrac{1}{3^{2017}}-\dfrac{1}{3}}{-\dfrac{2}{3}}\).