Ta có: (C₁): x²+ y² – 4 = 0 có tâm O (0; 0) và bán kính R= 2;

Dường tròn (C₂): (x-3)²+ (y-4) ²= 25 có tâm I( 3;4) và R= 5 nên OI= 5

Ta thấy: 5-2 < OI< 5+ 2

nên chúng cắt nhau.

Chọn B.

a) \(4x^2+y^2-25+4xy\\ =\left(4x^2+4xy+y^2\right)-25\\ =\left(2x+y\right)^2-5^2\\ =\left(2x+y-5\right)\left(2x+y+5\right)\)

b) \(\left(x-3\right)^2-\left(x+2\right)^2\\ =\left(x-3-x-2\right)\left(x-3+x+2\right)\\ =-5\left(2x-1\right)\)

a. 4x² + y² - 25 + 4xy

= (2x)² + 4xy - 5² - 5²

= (2x + 5)² - 5²

= (2x + 5 - 5)(2x + 5 + 5)

= 2x(2x + 10)

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

Đề bài yêu cầu gì vậy em.

d: \(x\left(x^2-1\right)+3\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)

e: \(x^2-10x+25=\left(x-5\right)^2\)

g: \(x^2-64=\left(x-8\right)\left(x+8\right)\)

h: \(\left(x+y\right)^2-\left(x^2-y^2\right)\)

\(=\left(x+y\right)\left(x+y-x+y\right)\)

\(=2y\left(x+y\right)\)

i: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

k: \(x^2+2xy+y^2-25=\left(x+y-5\right)\left(x+y+5\right)\)

l: \(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

a: \(5x-15y=5\left(x-3y\right)\)

b: \(5x^2y^2+15x^2y+30xy^2=5xy\left(xy+3x+6y\right)\)

c: \(x^3-2x^2y+xy^2-9x\)

\(=x\left(x^2-9-2xy+y^2\right)\)

\(=x\left(x-y-3\right)\left(x-y+3\right)\)

Đoạn từ sau chữ "Biết" thiếu dấu liên kết giữa $x_1,y_1,x_2,y_2$. Bạn cần viết lại đề rõ hơn.

Câu 8 A

Câu 7 C

Câu 6D

5D

4D

2C

1A

e camon nhiều

Bài 2:

a: \(7x^2-14xy+7y^2\)

\(=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\)

b: xy-3x+2y-6

=x(y-3)+2(y-3)

=(x+2)(y-3)

c: \(9x^2+6xy-25+y^2\)

\(=\left(9x^2+6xy+y^2\right)-25\)

\(=\left(3x+y\right)^2-25=\left(3x+y-5\right)\left(3x+y+5\right)\)

Bài 1:

a: 15x+10+4x(3x+2)=0

=>5(3x+2)+4x(3x+2)=0

=>(3x+2)(4x+5)=0

=>\(\left[\begin{array}{l}3x+2=0\\ 4x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac23\\ x=-\frac54\end{array}\right.\)

b: \(2x\left(x-6\right)+x^2-36=0\)

=>2x(x-6)+(x-6)(x+6)=0

=>(x-6)(2x+x+6)=0

=>(x-6)(3x+6)=0

=>(x+2)(x-6)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=6\end{array}\right.\)

Bài 1: Ta có 2009²⁰ = (2009²)¹⁰ = (2009.2009)¹⁰

                    20092009¹⁰ = (10001.2009)¹⁰

Mà 2009 < 10001 ➩ (2009.2009)¹⁰ < (10001.2009)¹⁰

Vậy 2009²⁰ < 20092009¹⁰

bài 2