1,Ta có :

x+y=7 =>\(\left(x+y\right)^2=7^2=49\)=> x^2+y^2+2xy=49

xy=12=> 2xy =24

=> x^2+y^2 +2xy-2xy =49-24=25=>x^2+y^2=25

=> x^2+y^2-2xy=25-24=1

=> (x-y)^2=1

=> Ix-yI=1

bài 2 mai giải tiếp nhé :))

`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.

`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`

`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`

`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.

Đặt \(\left\{{}\begin{matrix}x-y=a\\xy=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=7\\-ab=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=b+7\\ab+12=0\end{matrix}\right.\)

\(\Rightarrow\left(b+7\right)b+12=0\Leftrightarrow b^2+7b+12=0\Rightarrow\left[{}\begin{matrix}b=-3;a=4\\b=-4;a=3\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=4\\b=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=4\\xy=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+4\\xy+3=0\end{matrix}\right.\)

\(\Rightarrow\left(y+4\right)y+3=0\Rightarrow y^2+4y+3=0\Rightarrow\left[{}\begin{matrix}y=-1;x=3\\y=-3;x=1\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=3\\xy=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+3\\xy+4=0\end{matrix}\right.\)

\(\Rightarrow\left(y+3\right)y+4=0\Rightarrow y^2+3y+4=0\) (vô nghiệm)

Vậy hệ đã cho có 2 cặp nghiệm \(\left(x;y\right)=\left(3;-1\right);\left(1;-3\right)\)

Đặt \(\left\{{}\begin{matrix}x-y=a\\xy=b\end{matrix}\right.\) : Hệ trở thành ;

\(\left\{{}\begin{matrix}a-b=7\\ab=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\b^2+7b+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\\left(b+3\right)\left(b+4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=-3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\end{matrix}\right.\)

Với \(a=4;b=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\xy=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left(y+1\right)\left(y+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)

Với \(a=3;b=-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\xy=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\y^2+3y+4=0\end{matrix}\right.\) ( Vô nghiệm )

Vậy \(\left(x;y\right)=\left(3;-1\right)\) \(\left(x;y\right)=\left(1;-3\right)\)

Chứng minh gì bạn?

a/

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+x+y=20\)

\(\Leftrightarrow\left(x+y\right)^2+x+y-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=3\\x+y=-5\Rightarrow xy=12\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=12\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2+5t+12=0\left(vn\right)\)

b/

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=0\\\left(x+y\right)^2-2xy-x-y=22\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+2xy+2=0\\\left(x+y\right)^2-2xy-x-y-22=0\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+\left(x+y\right)-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=-5\\x+y=-5\Rightarrow xy=4\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\) thì x; y là nghiệm:

\(t^2-4t-5=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=5\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-1;5\right);\left(5;-1\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\) thì x; y là nghiệm:

\(t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-1;-4\right);\left(-4;-1\right)\)

c/

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=a\\y^2+y=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\) theo Viet đảo, a và b là nghiệm:

\(t^2-8t+12=0\Rightarrow\left[{}\begin{matrix}t=6\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=2\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x-6=0\\y^2+y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x-2=0\\y^2+y-6=0\end{matrix}\right.\end{matrix}\right.\)

Bạn tự bấm máy

a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)

\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)

\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)

\(Q=\left(x-y-2x-4y\right)^2\)

\(Q=\left(-x-5y\right)^2\)

b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)

\(A=\left[\left(xy+2\right)-2\right]^3\)

\(A=\left(xy+2-2\right)^3\)

\(A=\left(xy\right)^3\)

\(A=x^3y^3\)

c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)

\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)

\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)

\(=0\)

a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2

=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2

b: =(xy+2-2)^3=(xy)^3=x^3y^3

c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)

=24x+2x^3-2x^3-24x

=0

\(P=\left(x+y\right)\left\{\left[\left(x+y\right)^2-2xy\right]\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\right\}\\ \)

Thây số vào

VÌ \(x+y=7;xy=10\)

\(\Rightarrow x,y=5\)và \(2\)

\(\Rightarrow P=\left(5+2\right)\left(5^2+2^2\right)\left(5^3+2^3\right)\)

\(\Rightarrow P=7.29.133\)

    \(P=26999\)

26999

??