TL:

\(21-22x+y^2\)

\(=\left(y-11\right)^2\)

a. (2x-3)² = 36

(2x-3)² = 6²

=> TH1: 2x - 3 = 6

2x = 9

x = 9/2

TH2: 2x - 3 = -6

2x = -6 + 3

2x = -3

x = -3/2

Vậy x \(\in\){ -3/2 ; 9/2)

Câu b tương tự

a.(2x-3)^2=36

\(\Rightarrow\left(2x-3\right)^2=6^2\)

\(\Rightarrow2x-3=6\)

\(\Rightarrow2x=9\)\(\Rightarrow x=9:2=\frac{9}{2}\)

(x+1)^2>=0 và (y-1)^2>=0

=>C>=-10

Dấu = xảy ra khi x+1=0,y-1=0

=>x=-1,y=1

Vậy C=-10 khi x=-1,y=1

k cho mk nha

\(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{cases}\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2-10\ge-10}\)

Dấu ''='' xảy ra <=> x = -1 ; y = 1

A B C D O t x

có AOC và DOB đối đỉnh

\(\Rightarrow\widehat{AOC}=\widehat{DOB}\)

mà OX là tia phân giác AOC

\(\Rightarrow\widehat{AOT}=\widehat{TOC}\left(tc\right)\)

ta có\(\Rightarrow\widehat{AOC}=\widehat{DOB}\)(1)

mà ox là tia đối của  ot(2)

từ (1)và (2) =>ta ox là tia phân giác của DOB

Thank you bạn nha

Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)

Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)

Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)

\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)

\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)

\(2^{2010}-M=2^{2010}-1\)

=> M = 1

\(a-b=7\Leftrightarrow b=a-7\)

\(\Rightarrow P=\frac{3a-\left(a-7\right)}{2a-7}+\frac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)

\(=\frac{3a-a+7}{2a-7}+\frac{3a-21-a}{2a-14-7}\)

\(=\frac{2a+7}{2a-7}+\frac{2a-21}{2a-21}\)

\(=\frac{2a+7}{2a-7}+1=\frac{2a+7+2a-7}{2a-7}=\frac{4a}{2a-7}\)