Ta có:\(\frac{a}{b}=\frac{b}{c}=>\frac{a}{b}.\frac{b}{c}=\frac{a}{b}.\frac{a}{b}=>\frac{a.b}{b.c}=\frac{a^2}{b^2}=>\frac{a}{c}=\frac{a^2}{b^2}\)

\(\frac{a}{b}=\frac{b}{c}=>\frac{a}{b}.\frac{b}{c}=\frac{b}{c}.\frac{b}{c}=>\frac{a.b}{b.c}=\frac{b^2}{c^2}=>\frac{a}{c}=\frac{b^2}{c^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)

=>\(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

Bài 1:

a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2=x^2+2xy=x\left(x+2y\right)\)

b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)

\(=\left(x-y\right)^2\left(x+y\right)^2\)

c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)

\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)

\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)

Bài 2:

a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)

\(=2a\left(a^2+3b^2\right)\)

b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2b\left(b^2+3a^2\right)\)

a, \(\left(x+y\right)^2-y^2=x\left(x+2y\right)\Leftrightarrow x^2+2xy+y^2-y^2=x^2+2xy\)

\(\Leftrightarrow x^2+2xy=x^2+2xy\left(đpcm\right)\)

b, \(\left(x^2+y^2\right)-\left(2xy\right)^2=\left(x+y\right)^2\left(x-y\right)^2\)

\(\Leftrightarrow x^2+y^2-4x^2y^2=\left(x^2+2xy+y^2\right)\left(x^2-2xy+y^2\right)\)

\(\Leftrightarrow x^2+y^2-4x^2y^2=x^4-2x^2y^2+y^4\)đề sai ? 

\(A=3+3^2+3^3+...+3^{100}\)

\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(\Leftrightarrow2A=3^{101}-3\)

\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)

\(\Leftrightarrow A< B\)

a. tính A = 3+3^2+3^3+3^4+.....+3^100

3A=3^2+3^3+3^4+3^5+....+3^100

3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100

mà B=3^100-1 => A<B

\(A=1+4+4^2+...+4^{99}\)

\(\Leftrightarrow4A=4+4^2+4^3+...+4^{100}\)

\(\Leftrightarrow3A=4^{100}-1\)

\(\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)

hay A<B (đpcm)

 mk chẳng biết  nguyen hoang phi hung ak

chưa từng nhìn

botay'

huhi

câu 1:

số hữu tỉ đó là: 1,659842

số vô tỉ đó là: 1,58281134........

dễ mà

tôi chỉ thử để xem ai giỏi thì kb thôi

mà bài này tôi lấy trên mạng ấy mà