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Câu 1:
a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$
b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)
c.
\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)
d.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
\(\frac{7}{4}-x.\frac{3}{4}=\frac{5}{19}\)
\(\Rightarrow1+\frac{3}{4}-x.\frac{3}{4}=\frac{5}{9}\)
\(\Rightarrow\frac{3}{4}.\left(1-x\right)=\frac{5}{9}-1=-\frac{4}{9}\)
\(\Rightarrow1-x=-\frac{4}{9}:\frac{3}{4}=-\frac{16}{27}\)
\(\Rightarrow x=1-\left(-\frac{16}{27}\right)=1+\frac{16}{27}=1\frac{16}{27}\)
\(\frac{4}{7}\cdot\frac{2}{3}:x=\frac{4}{5}\)
(=)\(\frac{2}{3}x=\frac{4}{5}:\frac{4}{7}\)
(=)\(\frac{2}{3}x=\frac{7}{5}\)
(=) \(x=\frac{7}{5}:\frac{2}{3}\)
(=) \(x=\frac{21}{10}\)
a) 4/9.x=25/18:2/3
4/9.x=25/18.3/2
4/9.x=75/36
x=75/36:4/9
x=75/36x4/9
x=675/144
b)x:4/5=25/8:5/4
x:4/5=25/8x4/5
x:4/5=100/40
x:4/5=5/2
x=5/2x4/5
x=20/10
x=2/1=2
c)22/5:x=44/5:5/2
22/5:x=44/5x2/5
22/5:x=88/25
x=88/25:22/5
x=88/25x22/5
x=440/550
x=4/5
h nk
\(\frac{3}{4}.\frac{12}{7}+x=\frac{5}{2}\)
\(\frac{3.3}{1.7}+x=\frac{5}{2}\)
\(\frac{9}{7}+x=\frac{5}{2}\)
\(x=\frac{5}{2}-\frac{9}{7}\)
\(x=\frac{35}{14}-\frac{18}{14}\)
\(x=\frac{17}{14}\)
\(a,x-5⋮x+2\)
\(\Rightarrow x+2-7⋮x+2\)
\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
x + 2 = 1=> x = -1
x + 2 = -1 => x = -3
.... tương tự nhé ~
\(2x+3⋮x-5\)
\(\Rightarrow2x-10+7⋮x-5\)
\(\Rightarrow2\left(x-5\right)+7⋮x-5\)
\(\Rightarrow x-5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
x - 5 = 1 => x = 6
....
5^6+5^7+5^8
=5^6.(1+5+5^2)
=5^6.31 chia hết cho 31
7^6+7^5-7^4
=7^4.(7^2+7-1)
=7^4.55 chia hết cho 11
BÀI 2:
a) \(5^6+5^7+5^8=5^6\left(1+5+5^2\right)=5^6.31\) \(⋮\)\(31\)
b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\)\(⋮\)\(11\)
c) \(2^3+2^4+2^5=2^3.\left(1+2+2^2\right)=2^3.7\)\(⋮\)\(7\)
d) mk chỉnh đề
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{58}\right)\)
\(=3\left(1+2^2+...+2^{58}\right)\)\(⋮\)\(3\)
\(\dfrac{2}{5}\cdot x=\dfrac{4}{7}:\dfrac{2}{3}\)
\(\dfrac{2}{5}\cdot x=\dfrac{6}{7}\)
\(x=\dfrac{6}{7}:\dfrac{2}{5}\)
\(x=\dfrac{15}{7}\)
\(\dfrac{2}{5}.x=\dfrac{4}{7}:\dfrac{2}{3}\)
\(\dfrac{2}{5}.x=\dfrac{4}{7}.\dfrac{3}{2}\)
\(\dfrac{2}{5}.x=\dfrac{6}{7}\)
\(x=\dfrac{6}{7}:\dfrac{2}{5}\)
\(x=\dfrac{6}{7}.\dfrac{5}{2}=\dfrac{15}{7}\)