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2\(\frac{1}{2}\)x+x=2\(\frac{2}{15}\)\(\Rightarrow\)\(\frac{5}{2}\)x+x=\(\frac{32}{15}\)\(\Rightarrow\)x(\(\frac{5}{2}\)+1)=\(\frac{32}{15}\)\(\Rightarrow\)x.\(\frac{7}{2}\)=\(\frac{32}{15}\)\(\Rightarrow\)x=\(\frac{32}{15}\):\(\frac{7}{2}\)\(\Rightarrow\)x=\(\frac{32}{15}\).\(\frac{2}{7}\)\(\Rightarrow\)x=\(\frac{64}{105}\)
Giải
\(2\frac{1}{2}x-x=2\frac{2}{15}=>\frac{5}{2}x-x=\frac{32}{15}=>x\left(\frac{5}{2}-1\right)=\frac{32}{15}=>x.\frac{7}{2}=\frac{32}{15}=>x=\frac{32}{15}:\frac{7}{2}=>x=\frac{64}{105}\)
a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)
\(=-\frac{19}{8}\)
b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{5}{12}\)
c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)
\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)
\(=-\frac{43}{30}\)
d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)
\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)
\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)
\(=\frac{3}{20}\)
e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)
\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)
f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)
\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)
g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)
\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)
\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)
h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)
\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)
\(=\frac{78149}{6210}\)
i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)
\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)
\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)
\(=\frac{55}{4}\)
\(M=\frac{1}{2_{^{^2}}}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
=\(1-\frac{1}{n}<1\)
\(\Rightarrow M<1\)
1)
a)96-3(x+1)=42
3(x+1)=96-42
3(x+1)=54
x+1=54:3
x+1=18
x=18-1
x=17
b)12x-33=3^2.3^3
12x-33=243
12x=243+33
12x=276
x=276:12
x=23
c)24+5x=7^5:7^3
24+5x=49
5x=49-24
5x=25
x=25:5
x=5
d)(x+1)+(x+2)+.......+(x+100)=5750
100x+(1+2+...+100)=5750
100x+5050=5750
100x=5750-5050
100x=700
x=700:100
x=7
e)720:[41-(2x-5)]=2^3.5
720:[41-(2x-5)]=40
[41-(2x-5)]=720:40
[4
\(D=15+2^4+2^5+2^6+...+2^{2020}\\ 2D=30+2^5+2^6+2^7+...+2^{2021}\\ 2D-D=\left(30+2^5+2^6+2^7+...+2^{2021}\right)-\left(15+2^4+2^5+2^6+...+2^{2020}\right)\\ D=30+2^{2021}-15-2^4\\ D=2^{2021}+15-2^4=2^{2021}+15-16\\ D=2^{2021}-1\)
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