Trang Tran tra google đó nha.

Ta có:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow\)  \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow\)  \(a^2+b^2+c^2-\left(a^2+b^2+c^2\right)+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow\)  \(2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow\)  \(ab+bc+ac=0\)

Do đó   \(\frac{ab+bc+ac}{abc}=0\) , tức là  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Dễ dàng suy ra được   \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)  

Như vậy,  \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\left(đpcm\right)\)

The girl worked hard.Her stepmother was not happy.(Although)

Although the girl worked hard, her stepmother was not happy.

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giải bài sau

khối 3,4,5 có 288 hs

3<5     70 hs

5>4     20 hs

tìm 3,4,5

Tìm lỗi sai và sửa lại cho đúng : The toys keeping=>kept in the box are worth 30 dollars.

-> nO2 = 0.03 mol ; nAl2O3 = 0,02 mol

a. VO2= n.22,4= 0,03.22,4 = 0,672 l

ta có Vkk = 5VO2 = 5.0,672 = 3,36l

b ta có m Al2O3= n.M=0,02 . 102 =2,04g

a. K đối xứng với M qua I => I là trung điểm MK.
Xét tứ giác AKCM có: AC cắt MK tại I mà I là trung điểm của AC và MK => AKCM là hình bình hành.
Lại có góc AMC vuông (do tam giác amc cân nên trung tuyến am vuông góc với bc)
Vậy AKCM là hcn.
b. Do AKCM là hcn nên AK // và = MC hay AK // và = BM => tứ giác AKMB là hbh.
c. Để hcn AKCM là hình vuông thì AM = MC = 1/2 BC suy ra tam giác ABC có trung tuyến AM = 1/2 BC nên tam giác ABC vuông tại A. Vậy để AKCM là hình vuông thì tam giác ABC vuông cân tại A.

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