\(VT=x^2+2x\left(1-2y\right)+\left(1-2y\right)^2+\left(5y^2-\left(1-2y\right)^2-10y+14\right)\)

 \(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)  voi  moi  x;y

\(\Leftrightarrow2x^2=1-m^2-2m-2\)

\(2x^2=-\left(m+1\right)^2\)

\(VT\ge0;VP\le0\)

=> VT=VP=0

=> m=-1

khoong  ro days la gi

ta co  : a+b=CD-AB=30-9=21 (1)

pitago cho 2 tam giac vuong :  h²= 13²- a²= 20²- b²

^{=> (a+b)(b-a) = 231  (2)}

^{tu (1) va(2) =>  a=5; b=16}

^{=> h= 12}

^{S=(9+30).12:2=39.6=234}

tự vẽ hình nha.

vì ABCD là hình thang cân

=> DH=(CD- AB):2=(10-7):2=1,5

Mà AH/DH=4 => AH=4DH=4.1,5=6

S=(AB+CD).AH:2=(7+10).6:2=51 cm²

\(xy=2x\)

+ nếu x=0 =>  đúng với mọi y 

+ nếu x\(\ne0\) => y=2

vậy (x;y)=(0;y\(\in N\)); (x;y)=(x\(\in N^+;2\))

nửa chu vi là: 28:2=14m

d+r =14

mà d - r = 4

nên d=(14+4):2=9

 rộng r=9-4=5

S= d.r=9.5=45 m²

\(xy-2x=0\Leftrightarrow x\left(y-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0;mọi.y\\y=2;moi..x\end{array}\right.\)

\(\left(3x-2y\right)^2=9x^2-2.3x.2y+4y^2=9x^2-12xy+4y^2\)

\(=-\frac{2^{5xy}}{2^{4xy}}=-2^{xy}=-2^2=-4\)