a)\(\dfrac{\left(1,48+0.32\right).4,5}{0,25.4.20}\) .1,4 +4,33 b)1003,55 - 35,5.0,1-999 c)3.78.(200 -68)-3.78.(100-68) d) (1.5+1.8+...+4.5+4.8).0.1
giúp mink với ạ. Mình đang cần gấp
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20 tháng 2 2018
3,78 x ( 200 - 68 ) - 3,78 x ( 100 - 68 )
= 3,78 x 132 - 3, 78 x 32
= 3,78 x ( 132 - 32 )
= 3 , 78 x 100
= 378
15, 3 x 9,55 + 9,45 x 15 , 3 + 15 , 3
= 15,3 x ( 9.55 + 9.45 + 15.1 )
= 15,3 x 34.1
= 521.73
mk sợ câu thứ 2 chưa chắc đúng đâu nha câu nhờ ai giúp câu 2 nhé ^^
20 tháng 2 2018
a, \(3,78.\left(200-68\right)-3,78.\left(100-68\right)\)
\(=3,78.132-3,78.32\)
\(=3,78.\left(132-32\right)\)
\(=3,78.100\)
\(=378\)
b, \(15,3.9,55+9,45.15,3+15,3\)
\(=15,3.\left(9,55+9,45+1\right)\)
\(=15,3.20\)
\(=306\)
MV
1
12 tháng 4 2017
a) 3,78 x (200-68) - 3,78 x (100-68)
= 3,78 x 200 - 3,78 x 68 - 3,78 x 100 - 3,78 x 68
= 3,78 x (200-68-100-68)
= 3,78 x 100
= 378
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 12 2020
Đề bài có vấn đề, thay \(a=b=c\) hai vế cho kết quả khác nhau
Ta sẽ chứng minh BĐT mạnh hơn sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+3\sqrt{\dfrac{3\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(ab+bc+ca\right)^2}}=3+3\sqrt{Q}\)
Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)
\(\Rightarrow Q\le\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{abc}-3\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow\dfrac{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}{abc}\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\ge\dfrac{3}{\sqrt{2}}\sqrt{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(VT\ge3\sqrt[3]{abc\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\)
Do đó ta chỉ cần chứng minh:
\(8\left(abc\right)^2\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)^2\right]^2\ge\left(abc\right)^3\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow8\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\right]^2\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\dfrac{1}{8}\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^4\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) (hiển nhiên đúng theo AM-GM)
T
1
NA
1 tháng 3 2022
\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)
5 tháng 6 2022
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
7 tháng 6 2022
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
a: \(=\dfrac{1.8\cdot4.5}{0.25}\cdot\dfrac{1.4}{4.2}+4.33=\dfrac{32.4}{3}+4.33=10.8+4.33=15.13\)
b: \(=1003.55-3.55-999=1000-999=1\)
c: \(=3.78\left(200-68-100+68\right)=100\cdot3.78=378\)
d: \(=0.1\cdot\dfrac{\left(4.8+1.5\right)\cdot\left[\left(4.8-1.5\right):0.3+1\right]}{2}=3.78\)