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Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
a)= \(2^3\left(17-14\right)\)
\(=8.3\)
\(=24\)
b)\(=17\left(85+15\right)-120\)
\(=17.100-120\)
\(=1700-120\)
\(=1580\)
c)\(=\left(25.4\right).\left(125.8\right).\left(2.5\right)\)
\(=100.1000.10\)
\(=1000000\)
d)\(=24.53+24.87-24.40\)
\(=24\left(53+87-40\right)\)
\(=24.100\)
\(=2400\)
e)\(=5.7.77-7.60-7.7.25-15.6.7\)
\(=7\left(5.77-60-7.25-15.6\right)\)
\(=7\left(385-60-175-90\right)\)
\(=7.60\)
\(=420\)
\(15-\left(x-4\right)=8\)
\(x-4=15-8\)
\(x-4=7\)
\(x=7+4\)
\(x=11\)
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))
= \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\) x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)
= \(\dfrac{1}{100}\)
`@` `\text {Ans}`
`\downarrow`
`x^3 = 125`
`=> x^3 = 5^3`
`=> x = 5`
Vậy, `x = 5`
_____
`4^5 \div 4^3`
`=`\(4^{5-3}\)
`= 4^2`
______
\(8^{12}\cdot8^3\div8^{15}\)
`=`\(8^{12+3-15}\)
`=`\(8^0=1\)
______
\(7^{15}\div7^{13}\)
`=`\(7^{15-13}\)
`=`\(7^2\)
_______
\(17^{15}\cdot17^{10}\div17^{22}\)
`=`\(17^{15+10-22}\)
`= 17^3`
_______
\(5^{10}\cdot25\div5^{11}\)
`=`\(5^{10}\cdot5^2\div5^{11}\)
`=`\(5^{10+2-11}\)
`= 5`
_______
\(15^{10}\cdot225\div15^{11}\)
`=`\(15^{10}\cdot15^2\div15^{11}\)
`=`\(15^{12}\div15^{11}\)
`= 15`
`@` `\text {Kaizuu lv uuu}`
\(x^3=125\)
\(\Rightarrow x^3=5^3\)
\(\Rightarrow x=3\)
___________________
\(4^5:4^3\)
\(=4^{5-3}\)
\(=4^2=16\)
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\(8^{12}.8^3:8^{15}\)
\(=8^{15}:8^{15}\)
\(=1\)
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\(7^{15}:7^{13}\)
\(=7^{15-13}\)
\(=7^2=49\)
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\(17^{15}.17^{10}:17^{22}\)
\(=17^{25}:17^{22}\)
\(=17^3=4913\)
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\(5^{10}.25:5^{11}\)
\(=5^{10}.5^2:5^{11}\)
\(=5^{12}:5^{11}\)
\(=5^1=5\)
___________________
\(15^{10}.225:15^{11}\)
\(=15^{10}.15^2:15^{11}\)
\(=15^{12}:15^{11}\)
\(=15^1=15\)
__________________
* an . am = an+m
an : am = an-m
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