1. \(x^2+2y^2+2xy-2y+1=0\)

\(\left(x+y\right)^2+y^2-2y+1=0\)

\(\left(x+y\right)^2+\left(y-1\right)^2=0\)

Có: \(\left(x+y\right)^2\ge0;\left(y-1\right)^2\ge0\)

Mà theo bài ra: \(\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

Nhu lon

ko bit

a, 20^2 - 6^ 2=  400 - 36 = 364

b,3^3 . 18 - 3^3 . 12 = 3^3 . ( 18-12) = 27 . 6 = 162

c, 39 . 213 + 87 . 39 = ( 213 + 87) . 39 = 300 . 39 = 110700

d, 80 - [ 130 - ( 12 - 4)^2] = 80 - [130 - (3^2) ] = 80 - 130 - 9 = -59

Bài 1: a)  \(M=1+5+5^2+...+5^{100}\)

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

\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)

\(4M=5^{101}-1\)

\(M=\frac{5^{101}-1}{4}\)

b) \(N=2+2^2+...+2^{100}\)

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

\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(N=2^{101}-2\)

Bài 2:

a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\) 

\(32^{16}=\left(2^5\right)^{16}=2^{80}\)

Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)

MÌNH KHÔNG HIỂU GÌ CẢ

12 mũ 5 nhân 9 mũ 6 phần 27 mũ 5 nhân 2 mũ 12

25-27-(45:64 + 28:64)

-2 - ( 73:64)

a) 18¹⁰ : 3¹⁰ = (2.3²)¹⁰ : 3¹⁰ = 2¹⁰ . 3²⁰ : 3¹⁰ = 2¹⁰ . 3¹⁰ = 6¹⁰

b) 27⁶ : 9 = (3³)⁶ : 9 = 3¹⁸ : 3² = 3^18-2 = 3¹⁶

c) Thiếu dữ liệu đề bài

d) 2²⁵ : 32⁴ = 2²⁵ : (2⁵)⁴ = 2²⁵ : 2²⁰ = 2⁵

Bài 2 : 5.x² = 245 => x² = 245:5 = 49 => x = \(\pm\)7

a) 1 + 3 + 5 + ... + 13

= (13 + 1).[(13 - 1) : 2 + 1] : 2

= 14 . 7 : 2

= 49

= 7²

b) 3² + 4² + 12²

= 9 + 16 + 144

= 169

= 13²