\(A=\left(x+y\right)^2=\left(x-y\right)^2+4xy\)

Thay x - y = 5 và xy = 3 vào ta có:

\(5^2+4\cdot3=37\)

Vậy A = 37

B = \(\left(x-y\right)^2=\left(x+y\right)^2-4xy=7^2-4\cdot12=1\)

C sai đề?

D = \(x^2-y^2-2013x-2013y\)

\(=\left(x-y\right)\left(x+y\right)-2013\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(x+y\right)=0\)

E = \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x^2-y^2\right)\cdot0=0\)

1, https://hoc24.vn//hoi-dap/question/91350.html

Bài 3:

=>2xy-x-y-2=0

=>x(2y-1)-y+0,5-2,5=0

=>x(2y-1)-(y-0,5)=2,5

=>2x(2y-1)-(2y-1)=5

=>(2y-1)(2x-1)=5

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

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

Câu 5:

Đặt x/2=y/3=z/4=k

=>x=2k; y=3k; z=4k

x^2+y^2+z^2=116

=>4k^2+9k^2+16k^2=116

=>29k^2=116

=>k^2=4

TH1: k=2

=>x=4; y=6; z=8

TH2: k=-2

=>x=-4; y=-6; z=-8

Bài 1 : làm tương tự với bài 2;3 nhé

Ta có : \(f\left(0\right)=c=2010;f\left(1\right)=a+b+c=2011\)

\(\Rightarrow f\left(1\right)=a+b=1\)

\(f\left(-1\right)=a-b+c=2012\Rightarrow f\left(-1\right)=a-b=2\)

\(\Rightarrow a+b=1;a-b=2\Rightarrow2a=3\Leftrightarrow a=\dfrac{3}{2};b=\dfrac{3}{2}-2=-\dfrac{1}{2}\)

Vậy \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)

Câu 1:

k=x/y=-3

=>x=-3y

Câu 2:

k=xy=-4

=>x=-4/y

x+3 là bội của x-2

<=> x+3 chia hết cho x-2

<=> x-2+5 chia hết cho x-2

<=>(x-2)+5 chia hết cho x-2

<=> 5 chia hết cho x-2

=> x-2\(\in\)Ư(5)={-1,-5,1,5}

Vậy....

|x|=7; |y|=20 

TH1: x=7 ; y=20 <=> x-y=7-20=-13

TH2: x=-7; y=-20 <=> x-y=-7-(-20)=13

TH3: x=-7; y=20 <=> x-y=-7-20=-27

TH4: x=7; y=-20 <=> x-y=7--20=27

\(\left|x\right|\in3\) <=> x=-3 or x=3

\(\left|y\right|\in5\) <=> y=-5 or y=5

Để x-y=2 <=> x=3; y=5 sẽ thõa mãn điều kiện x-y=2

2) Ta có: \(\frac{x_1}{y_2}=\frac{x_2}{y_1}\Rightarrow\frac{x_1^2}{y_2^2}=\frac{x_2^2}{y_1^2}=\frac{x_1^2+x_2^2}{y_1^2+y_2^2}=\frac{2^2+3^2}{52}=\frac{1}{4}\)

\(\Rightarrow\frac{x_1^2}{y_2^2}=\frac{1}{4}\Rightarrow y_2^2=16\Rightarrow\)\(\orbr{\begin{cases}y_2=-4\\y_2=4\end{cases}\Rightarrow}\)\(\orbr{\begin{cases}y_1=-6\\y_1=6\end{cases}}\)

=> KL....

I2x+3I=x+2

TH1: Nếu \(x\le-\frac{3}{2}\)(*), =>I2x+3I=-2x-3

PT: -2x-3=x+2 <=> x=\(-\frac{5}{3}\)(tm (*))

TH2: Nếu \(x>-\frac{3}{2}\)(**), => I2x+3I=2x+3

PT: 2x+3=x+2 => x=-1 (tm (**))

Vậy x=...

3, \(\hept{\begin{cases}f\left(1\right)=a+b+c\\f\left(0\right)=a\\f\left(-1\right)=a-b+c\end{cases}}\)

Có: \(f\left(0\right)=c⋮3\)

\(f\left(1\right)+f\left(-1\right)=2a⋮3\Rightarrow a⋮3\)

\(f\left(1\right)-f\left(-1\right)=2b⋮3\Rightarrow b⋮3\)