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HM
27 tháng 4 2017
A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
HH
27 tháng 4 2017
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)
b)
\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)
23 tháng 9 2020
f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)
\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)
g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)
AF
23 tháng 9 2020
Bài giải
\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)
\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)
\(=2\)
\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)
8 tháng 4 2023
1:
a: \(A=2+3\sqrt{x^2+1}>=3\cdot1+2=5\)
Dấu = xảy ra khi x=0
b: \(B=\sqrt{x+8}-7>=-7\)
Dấu = xảy ra khi x=-8
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
21 tháng 3 2019
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
2 tháng 2 2022
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)
NT
1
VA
5
5 tháng 4 2022
a)
tại\(x = 1 , GTBT A(x)\) là:
\(5.1 ^3 − 3.1 + 4\)
\(= 5.1 − 3.1 + 4\)
\(= 5 − 3 + 4\)
\(= 2 + 4\)
\(=6\)
Vậy tại\(x = 1 , GTBT A ( x ) là 6\)
nhanh nghen
=\(\frac{7^3\times5^8}{7^2\times5^8}=7\)
xin tiick