a)\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2\cdot\left(x^3+1\right)+7x\cdot\left(x+1\right)=0\)

\(\Leftrightarrow2\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)+7x\cdot\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left[2\cdot\left(x^2+x+1\right)+7x\right]=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2-2x+2+7x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2+5x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(2x+1\right)\cdot\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)

b)\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)

\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\cdot\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=-66\)

ta có A=2+2^2+2^3+....+2^60

2A=2^2+2^3+2^4+....+2^61

2A-A=(2^2+2^3+2^4+....+2^61)-(2+2^2+2^3+....+2^60)

A=2^61-2

Vậy A<B do  2^61-2<2^61

Ta có : A=2+2^2+2^3+....+2^60

=> 2A = 2^2+2^3+....+2^61

=> 2A - A = 2^61 - 2

=> A = 2^61 - 2 < 2^61

Vậy A < B 

nhân 2 A rồi trừ đi A là ra

a=210                   

TA CÓ 1

1+2+3+....+20 CÓ 20-1+1 = 20 CẶP

TỔNG MỖI CẶP SỐ LÀ 20+1=21

CÓ TẤT CẢ 20 CẶP => 21 . 20 = 420

2:

a: A=1+2+2^2+2^3+2^4

=>2A=2+2^2+2^3+2^4+2^5

=>A=2^5-1

=>A=B

b: C=3+3^2+...+3^100

=>3C=3^2+3^3+...+3^101

=>2C=3^101-3

=>\(C=\dfrac{3^{101}-3}{2}\)

=>C=D

Ta có: 

\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)

\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)

 \(a.5^{27}=\left(5^3\right)^9=125^9\\ 2^{63}=\left(2^7\right)^9=128^9\)

Vì 128⁹ > 125⁹ => 2⁶³ > 5²⁷

^{\(5^{28}=\left(5^4\right)^7=625^7\\ 2^{63}=\left(2^9\right)^7=512^7\)}

Vì 625⁷ > 512⁷ = > 5²⁸ > 2⁶³

Đã CMR: \(5^{27}< 2^{63}< 5^{28}\)

\(b.A=1+2+2^2+2^3+2^4\\ 2A=2+2^2+2^3+2^4+2^5\\ 2A-A=\left(2+2^2+2^3+2^4+2^5\right)-\left(1+2+2^2+2^3+2^4+\right)\\ A=2^5-1\\ 2^5-1=2^5-1=>A=B\\ c,C=3+3^2+....+3^{100}\\ 3C=3^2+......+3^{101}\\ 3C-C=\left(3^2+...+3^{101}\right)-\left(3+...+3^{100}\right)\\ 2C=3^{101}-3\\ C=\dfrac{3^{101}-3}{2}\\ \dfrac{3^{101}-3}{2}=\dfrac{3^{101}-3}{2}=>C=D\)

\(B=\frac{31}{2}.\frac{32}{2}.....\frac{60}{2}\)

\(B=\left(31.32.33....60\right).\frac{1.2.3....60}{2^{30.\left(1.2.3...30\right)}}\)

\(B=\left(1.3.5.....59\right).\frac{2.4.6.....60}{2.4.6....60}=1.3.5...59\)

=> \(B=A\)

ta co:

2A=2(2 mu 60 +1 /2 mu 61 +1)

2A=2 mu 61 +2 / 2 mu 61 +1

2A=2 mu 61 +1+1/2 mu 61 +1

2A=1+1/2 mu 61 +1

ta co:

2B=2(2 mu 61 +1/2 mu 62 +1)

2B=2 mu 62 +2/2 mu 62+1

2B=2 mu 62 +1+1/2 mu 62 +1

2B=1+1/2 mu 62 +1

mà 1+1/2 mu 61+1>1+1/2 mu 62 +1 nen 2A >2B

vậy A>B

nhớ k đúng cho mk nha

Ta có:

2.A=2 mủ 61 +2/2 mủ 61 +1=1+(2/2 mủ 61 +1)

2.B=2 mủ 62 + 2 /2 mủ 62 +1=1+(2/2 mủ 62 + 1)

vì ... nên 2.A >2.B.Vậy A>B