ai giúp mik vs mik c.ơn
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H9
HT.Phong (9A5)
CTVHS
1 tháng 7 2023
Bài 5:
a) \(0,24\cdot-\dfrac{15}{4}=\dfrac{6}{25}\cdot-\dfrac{15}{4}=\dfrac{6\cdot-15}{25\cdot4}=-\dfrac{90}{100}=-\dfrac{9}{10}\)
b) \(4,5\cdot\dfrac{-4}{9}=\dfrac{9}{2}\cdot\dfrac{-4}{9}=\dfrac{9\cdot-4}{2\cdot9}=-\dfrac{4}{2}=-2\)
c) \(3,5\cdot-1\dfrac{2}{5}=\dfrac{7}{2}\cdot\dfrac{-7}{5}=\dfrac{7\cdot-7}{2\cdot5}=-\dfrac{49}{10}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 7 2023
Bài 6:
a) \(1\dfrac{1}{17}\cdot\left(-2\dfrac{1}{8}\right)=\dfrac{18}{17}\cdot\dfrac{-17}{8}=\dfrac{18\cdot-17}{17\cdot8}=\dfrac{-18}{8}=-\dfrac{9}{4}\)
b) \(\left(-2\dfrac{1}{3}\right)\cdot1\dfrac{1}{14}=-\dfrac{7}{3}\cdot\dfrac{15}{14}=\dfrac{-7\cdot15}{3\cdot14}=-\dfrac{5}{2}\)
c) \(1,25\cdot\left(-3\dfrac{3}{8}\right)=\dfrac{5}{4}\cdot-\dfrac{27}{8}=\dfrac{5\cdot-27}{4\cdot8}=-\dfrac{135}{32}\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 7 2023
Bài 3:
a) \(\left(-\dfrac{2}{3}\right)^2\cdot\left(\dfrac{2}{3}\right)^5\)
\(=\left(\dfrac{2}{3}\right)^2\cdot\left(\dfrac{2}{3}\right)^5\)
\(=\left(\dfrac{2}{3}\right)^{2+5}\)
\(=\left(\dfrac{2}{3}\right)^7\)
b) \(\left(-\dfrac{1}{2}\right)^5\cdot\left(-\dfrac{1}{2}\right)^3\)
\(=\left(-\dfrac{1}{2}\right)^{5+3}\)
\(=\left(-\dfrac{1}{2}\right)^8\)
\(=\left(\dfrac{1}{2}\right)^8\)
c) \(\left(\dfrac{6}{5}\right)^7\cdot\left(-\dfrac{6}{5}\right)^4\)
\(=\left(\dfrac{6}{5}\right)^7\cdot\left(\dfrac{6}{5}\right)^4\)
\(=\left(\dfrac{6}{5}\right)^{7+4}\)
\(=\left(\dfrac{6}{5}\right)^{11}\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 7 2023
Bài 4:
a) \(\left(\dfrac{3}{7}\right)^4:\left(-\dfrac{3}{7}\right)^2\)
\(=\left(\dfrac{3}{7}\right)^4\cdot\left(\dfrac{3}{7}\right)^2\)
\(=\left(\dfrac{3}{7}\right)^{4+2}\)
\(=\left(\dfrac{3}{7}\right)^6\)
b) \(\left(\dfrac{5}{9}\right)^{11}:\left(\dfrac{5}{9}\right)^7\)
\(=\left(\dfrac{5}{9}\right)^{11-7}\)
\(=\left(\dfrac{5}{9}\right)^4\)
c) \(\left(\dfrac{2}{13}\right)^7:\left(\dfrac{2}{13}\right)^5\)
\(=\left(\dfrac{2}{13}\right)^{7-5}\)
\(=\left(\dfrac{2}{13}\right)^2\)
R
1
RY
24 tháng 4 2022
a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\)
= \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)
= \(\dfrac{2}{2}+\dfrac{3}{2}-\dfrac{4}{2}\)
= \(\dfrac{1}{2}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 11 2023
\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)
\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)
\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)
\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)
\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)
\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)
Ta có: 5 ⋮ 5
⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 11 2023
A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78
A = (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)
A = 7.(1 + 72) + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)
A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)
A = 7.50 + 72.50 + 75.50 + 76.50
A = 50.(7 + 72 + 75 + 76)
Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm
20 tháng 3 2021
a) Ta có: \(\widehat{xOt}+\widehat{tOy}=90^0\)(hai góc phụ nhau)
\(\Leftrightarrow\widehat{tOy}=90^0-\widehat{xOt}=90^0-40^0\)
hay \(\widehat{tOy}=50^0\)
Vậy: \(\widehat{tOy}=50^0\)
Bài 3:
Số tiền An góp chiếm:
1-1/5-1/5-1/7=16/35(tổng số tiền)
Số tiền quả bóng là:
30500:16/35=66718,75(đồng)