Này cậu ơi!

áp dụng bất đẳng thức cô si ta có:

\(a^2+b^2\ge2ab\Leftrightarrow2\left(a^2-ab+b^2\right)\ge a^2+b^2\)

\(\Leftrightarrow2\left(a^3+b^3\right)\ge\left(a+b\right)\left(a^2+b^2\right)\)

\(\Leftrightarrow\frac{a^3+b^3}{2}\ge\frac{a+b}{2}.\frac{a^2+b^2}{2}\)

A=2-5+8-11+...+98-101

=(2-5)+(8-11)+...+(98-101)

=(-3)+(-3)+...+(-3)(17 số hạng -3)

=(-3).17

=-51

thiet ko

24 mắt

A B C D E 1 1 1 2 2 1

\(\Delta ABC\)cân tại A nên\(\widehat{ABC}=\widehat{ACB}=\frac{180^0-\widehat{BAC}}{2}=75^0\)

Trên nửa mặt phẳng bờ BC chứa A lấy E sao cho\(\widehat{B_1}=\widehat{C_1}=45^0\)

=>\(\widehat{ABE}=75^0-45^0=30^0;\Delta EBC\)vuông cân tại E =>\(BE=EC=\frac{BC}{\sqrt{2}}=\sqrt{2}\left(cm\right)\)(định lí Pitago)

\(\Delta ABE,\Delta BAD\)có AB chung ; BE = AD\(\left(=\sqrt{2}cm\right)\);\(\widehat{ABE}=\widehat{BAD}\left(=30^0\right)\)

\(\Rightarrow\Delta ABE=\Delta BAD\left(c.g.c\right)\Rightarrow\widehat{A_1}=\widehat{B_2}\)

Lại có\(\Delta AEB=\Delta AEC\left(c.c.c\right)\)nên\(\widehat{A_1}=\widehat{A_2}=\frac{\widehat{BAC}}{2}=15^0\Rightarrow\widehat{B_2}=15^0\)

\(\Rightarrow\widehat{D_1}=\widehat{BAD}+\widehat{B_2}=45^0\)(\(\widehat{D_1}\)là góc ngoài\(\Delta ABD\)) ;\(\widehat{DBC}=75^0-15^0=60^0\)

\(\Delta BDC\)có\(\widehat{D_1}< \widehat{DBC}< \widehat{DCB}\left(45^0< 60^0< 75^0\right)\)nên BC < DC < BD

bai nay trong sach nang cao toan 7 trang 141

44100 đó bạn

Nhớ k nhé

Ta có công thức :

Với mọi n thuộc N thì :

\(1^2+2^2+3^2+.......+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Áp dụng vào bài toán ta được :

\(A=1^2+2^2+3^2+....+20^2=\frac{20\left(20+1\right)\left(2.20+1\right)}{6}=2870\)

|x+23|^2002|+|y-12|^123=0

<=>|x+23|^2002\(\ge\) 0

      \(\left|y-12\right|^{^{123}}\ge0\)

=>\(\left|x+23\right|^{2002}+\left|y-12\right|^{123}\ge0\)

=>x+23=0=>x=-23

=>y-12=0=>y=12

\(\left|x+23\right|^{2002}+\left|y-12\right|^{123}=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+23\right|^{2002}=0\\\left|y-12\right|^{123}=0\end{cases}}\Rightarrow\hept{\begin{cases}x+23=0\\y-12=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-23\\y=12\end{cases}}\)

Vậy \(x=-23;y=12\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+........+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Chúc bạn học tốt!!!!!!

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

A=1.2+2.3+3.4+....+n(n+1)

3a=1.2.(3-0)+2.3(4-1)...+n(n+1)[n+2-n+1]

3a=1.2.3-1.2.0+2.3.4-2.3.1+.....+n(n+1).(n+2)-n(n+1).n+1

3a=n(n+1).(n+2)

a=n(n+1)(n+2)/3

Đặt \(A=1\cdot2+2\cdot3+3\cdot4+..................+99\cdot100\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...........+99\cdot100\cdot\left(101-98\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+..............+99\cdot100\cdot101-98\cdot99\cdot100\)

\(3A=99\cdot100\cdot101\)

\(\Rightarrow A=\frac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=333300\)

1.2+2.3+3.4+...+99.100=?

1.2 + 99.100 = 100.5 x 50 (cặp)=5015 

theo mình là vậy !