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

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

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

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

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

\(\Leftrightarrow A< B\)

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)

Ta có:

a//b và a//c

⇒a⊥b và a⊥c

vì 1 đường thẳng cắt 2 đường thẳng và vuông góc với cả 2 thì 2 đường thẳng còn lại song song với nhau

⇒b//c

a)a²+b²+c²+3=2(a+b+c)

=>a²+b²+c²+1+1+1-2a-2b-2c=0

=>(a²-2a+1)+(b²-2b+1)+(c²-2c+1)=0

=>(a-1)²+(b-1)²+(c-1)²=0

=>a-1=b-1=c-1=0 <=>a=b=c=1 

-->Đpcm

b)(a+b+c)²=3(ab+ac+bc)

=>a²+b²+c²+2ab+2ac+2bc -3ab-3ac-3bc=0 

=>a²+b²+c²-ab-ac-bc=0

=>2a²+2b²+2c²-2ab-2ac-2bc=0 

=>(a²- 2ab+b²)+(b²-2bc+c²) + (c²-2ca+a²) = 0

=>(a-b)²+(b-c)²+(c-a)²=0 

Hay (a-b)²=0 hoặc (b-c)²=0 hoặc (a-c)²=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

c)a²+b²+c²=ab+bc+ca

=>2(a²+b²+c²)=2(ab+bc+ca)

=>2a²+2b²+c²=2ab+2bc+2ca

=>2a²+2b²+c²-2ab-2bc-2ca=0

=>a²+a²+b²+b²+c²+c²-2ab-2bc-2ca=0

=>(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ca+c²)=0

=>(a-b)²+(b-c)²+(a-c)²=0

Hay (a-b)²=0 hoặc (b-c)²=0 hoặc (a-c)²=0

=>a-b hoặc b=c hoặc a=c

=>a=b=c 

-->Đpcm

a) Ta có : \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Vì \(\left(a-1\right)^2\ge0,\left(b-1\right)^2\ge0,\left(c-1\right)^2\ge0\) nên pt trên tương đương với \(\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\) \(\Leftrightarrow a=b=c=1\)

b) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\) (1)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ac\)

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

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

Mà \(\left(a-b\right)^2\ge0,\left(b-c\right)^2\ge0,\left(c-a\right)^2\ge0\)

\(\Rightarrow\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\) \(\Rightarrow a=b=c\)

c) Giải tương tự câu b) , bắt đầu từ (1)

\(\left(a-b\right)^3+3ab\left(a-b\right)=a^3-b^3\)

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

\(\Leftrightarrow a^3-b^3-3ab\left(a-b\right)+3ab\left(a-b\right)=a^3-b^3\)

\(\Leftrightarrow a^3-b^3=a^3-b^3\) (luôn đúng)

1a)Xét a² + 5 - 4a =a² - 4a + 4+1=(a - 2)²+1\(\ge\)1 hay (a -2)² + 1 > 0 

\(\Rightarrow\)Đpcm

  b)Xét 3(a² + b² + c²) -(a + b +c)² =3a² + 3b² + 3c² - a² - b² - c² - 2ab - 2ac - 2bc

                                                  =2a² + 2b² + 2c²  - 2ab - 2ac - 2bc

                                                  =(a - b)² + (a - c)² + (b - c)²\(\ge\)0 (với mọi a,b,c)

\(\Rightarrow\)Đpcm

2)Xét A=\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+c+b\right)=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

         áp dụng cô-sy

\(\Rightarrow\)A\(\ge\)9

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)

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

`(a-b)^3 = [-(b-a)]^3 = (-1)^3 . (b-a)^3=-(b-a)^3`