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Ta đặt tổng sau là \(A=1^2+2^2+3^2+...+100^2\)
\(A=1+2.\left(1+1\right)+3.\left(2+1\right)+..+100\left(99+1\right)\)
\(A=1+2.1+2+3.2+3+..+100.99+100\)
\(A=\left(1.2+2.3+...+100.99\right)+\left(1+2+3+..+100\right)\)
\(A=333000+5050\)
\(A=338050\)
Vậy \(A=338050\)
a) \(M=\left\{0;2;4;6;8\right\}\)
b)\(N=\left\{1;3;5;7;9\right\}\)
c)\(P=\left\{10;15;20;25;30\right\}\)
d)\(H=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
\(I=\left\{51;54;57\right\}\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1
Ta có :
\(x^2\ge0\)
\(\Leftrightarrow\)\(3x^2\ge0\)
\(\Leftrightarrow\)\(3x^2+2016\ge2016\)
\(\Leftrightarrow\)\(\frac{3}{3x^2+2016}\le\frac{3}{2016}\)
\(\Leftrightarrow\)\(\frac{3}{3x^2+2016}\le\frac{1}{672}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x^2=0\)
\(\Leftrightarrow\)\(x=0\)
Vậy GTLN của \(A\) là \(\frac{1}{672}\) khi \(x=0\)
Chúc bạn học tốt ~
Ta có :
\(\left|x-2018\right|\ge0\)
\(\Leftrightarrow\)\(2\left|x-2018\right|\ge0\)
\(\Leftrightarrow\)\(2\left|x-2018\right|+3\ge3\)
\(\Leftrightarrow\)\(\frac{1}{2\left|x-2018\right|+3}\le\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|x-2018\right|=0\)
\(\Leftrightarrow\)\(x-2018=0\)
\(\Leftrightarrow\)\(x=2018\)
Vậy GTLN của \(B\) là \(\frac{1}{3}\) khi \(x=2018\)
Chúc bạn học tốt ~
a) 7/13.7/15 - 5/12.21/39 + 49/91.8/15
= 7/13. 7/15 - 5/12. 7/13 + 7/13.8/15
= 7/13. ( 7/15 - 5/12 + 8/15)
= 7/13. ( 7/15 + 8/15 - 5/12)
= 7/13. ( 1 - 5/12)
= 7/13. 7/12
= 49/156
b) ( 12/199 + 23/100 - 34/201) . ( 1/2-1/3-1/6)
= ( 12/199 + 23/100 - 34/201).0
= 0
a) \(=\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{130}.\frac{8}{15}=\frac{7}{13}\left(\frac{7}{15}+\frac{8}{15}-\frac{5}{12}\right)=\frac{7}{13}\left(1-\frac{5}{12}\right)=\frac{7}{13}.\frac{7}{12}=\frac{48}{156}\)
b) \(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0=0\)