a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-11;1\right)\right\}\)

ủa lớp 1 đâu có học cái này 

a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)

a: Phương trình hoành độ giao điểm là:

\(-x^2=\left(3-m\right)x+2-2m\)

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

\(\Delta=\left(3-m\right)^2-4\cdot1\cdot\left(2-2m\right)\)

\(=m^2-6m+9-8+8m=m^2+2m+1=\left(m+1\right)^2>0\forall m<>-1\)

=>(P) luôn cắt (d) tại hai điểm phân biệt

b: Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=-\left(3-m\right)=m-3;x_1x_2=\frac{c}{a}=2-2m\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=\left(m-3\right)^2-4\left(2-2m\right)=m^2-6m+9-8+8m=m^2+2m+1=\left(m+1\right)^2\)

\(\left(x_1+x_2\right)^2=\left(m-3\right)^2\)

\(\left|y_{A}-y_{B}\right|=2\)

=>\(\left|-x_1^2+x_2^2\right|=2\)

=>\(\left|x_1^2-x_2^2\right|=2\)

=>\(\left(x_1+x_2\right)^2\cdot\left(x_1-x_2\right)^2=4\)

=>\(\left\lbrack\left(m+1\right)\left(m-3\right)\right\rbrack^2=4\)

=>\(\left\lbrack m^2-2m-3\right\rbrack^2=4\)

=>\(\left[\begin{array}{l}m^2-2m-3=2\\ m^2-2m-3=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}m^2-2m-5=0\\ m^2-2m-1=0\end{array}\right.\)

TH1: \(m^2-2m-5=0\)

=>\(m^2-2m+1-6=0\)

=>\(\left(m-1\right)^2=6\)

=>\(\left[\begin{array}{l}m-1=\sqrt6\\ m-1=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}m=\sqrt6+1\\ m=-\sqrt6+1\end{array}\right.\)

TH2: \(m^2-2m-1=0\)

=>\(m^2-2m+1-2=0\)

=>\(\left(m-1\right)^2=2\)

=>\(\left[\begin{array}{l}m-1=\sqrt2\\ m-1=-\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}m=\sqrt2+1\\ m=-\sqrt2+1\end{array}\right.\)

Đáp án D

Xét phương trình hoành độ giao điểm

x² – 4x – 1 = 0 

Giả sử A(2 +  5 ;  5 ); B(2 -  5 ; - 5 ) => y_A + y_B = 0

Đáp án D

Phương trình hoành độ giao điểm 

x 3 − x + 3 = 2 x + 1 ⇔ x 3 − 3 x + 2 = 0

⇔ x − 1 2 x + 2 = 0 ⇔ x = − 2 x = 1

Do x A > x B  nên x A = 1, x B = − 2 → y B = 2 − 2 + 1 = − 3

Vậy  x B + y B = − 2 + − 3 = − 5

\(a.\)

\(\dfrac{x}{2}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)

\(b.\)

\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)

\(c.\)

\(\dfrac{x}{4}=\dfrac{y}{5}\)

\(\Leftrightarrow\dfrac{2x}{8}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

mà x+y=35

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(10;25)

b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

mà y-3x=2 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy: (x,y)=(1;5)

c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)

nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)

mà 2x-y=15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(20;25)

Giải:

a) \(y^2=3-\left|2x-3\right|\) 

Vì \(-\left|2x-3\right|\le0\forall x\) nên \(3-\left|2x-3\right|\le3\forall x\) nên \(y^2\le3\rightarrow y^2\in\left\{0;1\right\}\) (vì \(y\in Z\) )

TH1:

\(y^2=0\) 

\(\Rightarrow y=0\) 

\(\Rightarrow\left|2x-3\right|=3\) 

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) 

TH2:

\(y^2=1\) 

\(\Rightarrow y=\pm1\)

ơ xin lỗi lỡ nhấn gửi

a: xy=x-y

=>xy-x+y=0

=>xy-x+y-1=-1

=>x(y-1)+(y-1)=-1

=>(x+1)(y-1)=-1

=>\(\left(x+1\right)\left(y-1\right)=1\cdot\left(-1\right)=\left(-1\right)\cdot1\)

=>\(\left(x+1;y-1\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;0\right);\left(-2;2\right)\right\}\)

b: x(y+2)+y=1

=>\(x\left(y+2\right)+y+2=3\)

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

=>\(\left(x+1\right)\cdot\left(y+2\right)=1\cdot3=3\cdot1=\left(-1\right)\left(-3\right)=\left(-3\right)\left(-1\right)\)

=>\(\left(x+1;y+2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;1\right);\left(2;-1\right);\left(-2;-5\right);\left(-4;-3\right)\right\}\)

giúp mình với, mình đang vội!

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\)\(\dfrac{x}{3}\times\dfrac{y}{7}=\dfrac{xy}{21}=\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{7}\right)^2\)

\(\dfrac{xy}{21}=\dfrac{84}{21}=4\)

\(\Rightarrow\left(\dfrac{x}{3}\right)^2=4\Rightarrow\)\(\dfrac{x}{3}=2\Rightarrow x=6\)

\(\Rightarrow\left(\dfrac{y}{7}\right)^2=4\Rightarrow\)\(\dfrac{y}{7}=2\Rightarrow y=14\)