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\(\frac{2}{5}:\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{6}{5}+\frac{-2}{3}+\frac{1}{5}\)
\(=\frac{11}{15}\)
~ Hok tốt ~
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=4.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2008.2010}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=4.\left[\frac{1}{2}+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{2008}-\frac{1}{2008}\right)-\frac{1}{2010}\right]\)
\(=4.\left[\frac{1}{2}-\frac{1}{2010}\right]\)
\(=4.\frac{502}{1005}=\frac{2008}{1005}\)
~ Hok tốt ~
\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b,\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
Vậy
a, \(A=2015.20162016-2016.20152015\)
\(A=2015.\left(2016.10001\right)-2016.20152015\)
\(A=\left(2015.10001\right).2016-20152015.2016\)
\(A=20152015.2016-20152015.2016\)
\(A=0\)
Vậy A = 0
b, \(B=\left(3.4.2^{16}\right)^2\div11.2^{13}.4^{11}-16^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.2^{22}-2^{36}\)
\(B=3^2.2^{36}\div11.2^{35}-2^{36}\)
\(B=3^2.2^{35}.2\div11.2^{35}-2.2^{35}\)
\(B=3^2.2\div9=9.2\div9=2\)
Vậy B = 2
c, \(C=2^{10}.13+2^{10}.65\div2^8.104\)
\(C=2^{10}.\left(13+65\right)\div2^8.104\)
\(C=2^{10}.78\div2^8.104\)
\(C=2^{10}.39\div2^8.13\)
\(C=39\div13=3\)
Vậy C = 3
Đề bài câu c sai mk sửa nhé là 28 ms tính đc k nó dư lắm !!!
1)thay x-4 vào bt ta có:
(-25).4.(-3)
=(-100).(-3)
=300
2)thayy=25 vào bt ta có:
=(-1).[(-4).25].(5.8)
=(-1).(-100).40
=4000
3/ thay a = 4; b = -6; c = 12 vào bt ta có:
(2.4(-6)2) : 12=(2.144)-12=288:12=24
4/thay x = 4; y = -9 vào bt ta có:
[(-25).(-27).(-4)] : (-9)
=-2500:(-9)
=-\(\frac{2500}{9}\)
5/ thay a = 5 ; b = -3 vào bt ta có:
[52 - (-3)2] : [5 + (-3)][5 – (-3)]
=[25-9]:2.8
=16:2.8
=64
a: \(=5^2+2^5-1=25+32-1=25+31=56\)
c: \(=\dfrac{-7}{25}\cdot\dfrac{39}{14}\cdot\dfrac{50}{78}\)
\(=\dfrac{-7}{14}\cdot\dfrac{39}{78}\cdot\dfrac{50}{25}=\dfrac{-1}{2}\)
d: \(=\left(\dfrac{3}{8}-\dfrac{3}{4}+\dfrac{7}{12}\right)\cdot\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{9-18+14}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)
a: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2-10+3}{15}=\dfrac{-5}{15}=\dfrac{-1}{3}\)
b: \(=\left(6+\dfrac{1}{8}-\dfrac{1}{2}\right)\cdot4=\dfrac{48+1-4}{8}\cdot4=\dfrac{45}{2}\)
c: \(=\dfrac{1}{4}\cdot4-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
d: \(F=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2008\cdot2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{1004}{1005}\)