Đặt \(A=1+2+2^2+...+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}+2^{101}\)

\(\Rightarrow2A-A=-1+2^{101}\)

\(\Rightarrow A=2^{101}-1\)

Đặt A = 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰

2A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹

2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹) - (1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰)

A = 2¹⁰¹ - 1 (đpcm)

Đặt A = 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰

2A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹

2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹) - (1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰)

A = 2¹⁰¹ - 1 (đpcm)

đpcm là gì vậy

VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

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

\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)

\(=10-1=9=VP\)

gọi biểu thức trên là A. Ta có:

\(2A=2\left(1^2+...+1^{100}\right)\)

\(=2+2^2+...+2^{101}\)

\(2A-A=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)

\(A=2^{101}-1\)

Bạn xem lại đề

dùng hàng đẳng thức A^2-B^2=(A-B)(A+B) nhé còn phần b chuyển vế sang rồi dùng HĐT là được

a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

b) \(100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2\)

\(\Leftrightarrow\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105-107^2\right)+\left(94^2-96^2\right)=0\)

\(\Leftrightarrow2\left(100+98+103+101-105-107-94-96\right)=0\)

\(\Leftrightarrow2\times0=0\)(ĐPCM)

Xét hiệu :

\(100^2+103^2+105^2+94^2-\left(101^2+98^2+96^2+107^2\right)\)

\(=100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)

\(=\left(100^2-98^2\right)+\left(103^2+101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)

\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(96-94\right)\left(96+94\right)\)\(-\left(107-105\right)\left(107+105\right)\)

\(=2.198+2.204-2.212-2.190\)

\(=2.\left(198+204-212-190\right)\)

\(=2.0\)

\(=0\)

VẬY dpcm

Ta có:  

100²+103²+105²+94²=101²+98²+96²+107²

=>100²+103²+105²+94²-(101²+98²+96²+107²)=0

=>100²+103²+105²+94²-101²-98²-96²-107²=0

=>(100²-98²) + (103²-101²) + (105²-107²) + (94²-96²) = 0

=>(100-98)(100+98) + (103-101)(103+101) + (105-107)(105+107) + (94-96)(94+96) = 0

=>2.(100+98) + 2.(103+101) - 2.(105+107) - 2.(94+96) = 0

=>2.[(100+98)+(103+101)-(105+107)-(94+96)] = 0

=>2.(198+204-212-190)=0

=>2.0=0

                     Chứng tỏ 100²+103²+105²+94²=101²+98²+96²+107²

cám ơn các bn nhiều ạ!!!

3²¹>2³¹

3^{21 _{> 2}31}

\(3^{21}=\left(3^3\right)^7=27^7\)

\(2^{31}=\left(2^4\right)^7.2=\left(2^5\right)^7=32^7\)

Vì 27 < 32 nên \(3^{21}< 2^{31}\)