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 ^^

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, \(15,3.9,55+9,45.15,3+15,3\)

\(=15,3.\left(9,55+9,45+1\right)\)

\(=15,3.20\)

\(=306\)

Kết Quả là 37.8

=1658,40327

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

a) 3,78 x (200 - 68) - 3,78 x (100 - 68)

= 3,78 x [(200 - 68) - (100 - 68)]

= 3, 78 x [200 - 68 -100 +68 ]

= 3,78 x 100 =378

b> 1,5 + 1,8 + 2,1........ + 4,5 + 4,8 

Nhận xét: 1,8 -1,5 = 0,3 

               2,1 - 1,8 = 0,3 

            ........................

            4,8 - 4,5 = 0,3 

Quy luật : Mỗi số hạng liền kề hơn kém nhau 0,3 đơn vị 

                Số các số hạng là :

                        ( 4,8 - 4,5 ) :0,3 + 1= 12 ( số ) 

                Tổng của dãy số là : 

                           ( 4,8 + 4,5 ) x 12 : 2 = 37,8

                                           Đáp số : 37,8 

 CHÚC MN HC TỐT NHA >< :3

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)

\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)

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: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

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: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)