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a, \(\dfrac{15\times14-1}{13\times15+14}\)
= \(\dfrac{15\times\left(13+1\right)-1}{13\times15+14}\)
= \(\dfrac{13\times15+15-1}{13\times15+14}\)
= \(\dfrac{13\times15+14}{13\times15+14}\)
= 1
c, \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)
= \(\dfrac{\left(9+1\right)\times\left\{\left(9-1\right):1+1\right\}:2}{10}\)
= \(\dfrac{10\times9:2}{10}\)
= \(\dfrac{9}{2}\)
\(a,2017\times\left(8+2\right)=2017\times10=20170\\ b,=\dfrac{3}{7}\times\left(\dfrac{9}{7}-\dfrac{2}{7}\right)=\dfrac{3}{7}\times1=\dfrac{3}{7}\)
Lời giải:
$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$
$=\underbrace{2+2+2+...+2}_{24}+7$
$=2\times 24+7$
$=55$
Ta có: 99–97 +95–93+...+7–5+3−1
=(99−97)+(95−93)+...+(7−5)+(3−1)
=2+2+...+2+2 (có 25 số 2)
=2.25
=50
số số hạng là :`(26 - 2) : 2 + 1 = 13(số hạng)`
`13 : 2 = 12 dư 1 `
`=>A = (2 - 4) + (6 - 8) + (10 - 12) + ... + (22 - 24) + 26`
`=>A = (-2) + (-2) + (-2) + ... + (-2) + 26`
`=>A = (-2) xx 6 + 26`
`=> A = -12 + 26`
`=> A = 14`
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
2:
b=2000*2004
=(2002-2)*(2002+2)
=2002^2-4
=>b<a
1:
a: \(=8\cdot9\left(14+17+19\right)=72\cdot50=3600\)
Bài 1:
\(8\times9\times14+6\times17\times12+19\times4\times18\)
\(=8\times9\times14+3\times2\times17\times2\times2\times3+19\times4\times2\times9\)
\(=8\times9\times14+17\times8\times9+19\times8\times9\)
\(=8\times9\times\left(14+17+19\right)\)
\(=8\times9\times50\)
\(=72\times5\times10\)
\(=360\times10\)
\(=3600\)
Bài 2:
Ta có:
\(a=2022\times2022\)
Và: \(b=2000\times2004\)
Mà: \(2022>2000,2022>2004\)
\(\Rightarrow2022\times2022>2000\times2004\)
\(\Rightarrow a>b\)
1
a)48,124
b)-28,538
c)56,96
d)111,5
Câu 2
a)132,1
b)21,35
Câu 3
a)4,8
b)71,5
