ai giải giúp em đi ạ em đang cần gấp lắm ạ 

Cả hai baif hộ mik nhé

x^2 - x - y^2 - y 

= x^2 - y^2 - x - y 

= ( x - y ) ( x + y ) - ( x + y ) 

= ( x + y ) ( x - y - 1  ) 

x^2 - 2xy + y^2 - z^2 

= ( x- y ) ^2 - z^2 

= ( x - y - z ) ( x - y + z ) 

Bài 9. 

d) \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\).

Bài 10.

\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2-2x+1+y^2+4y+4=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\).

Trả lời:

Bài 1:

a, \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)\)

\(=8x^3+36x^2+54x+27+8x^3-36x^2+54x-27-8x^2+18\)

\(=16x^3-8x^2+108x+18\)

b, \(\left(x+2\right)^3+\left(x-2\right)^3+x^3-3x\left(x+2\right)\left(x-2\right)\)

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8+x^3-3x\left(x^2-4\right)\)

\(=3x^3+24x-3x^3+12x=36x\)

Bài 2:

a, \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)

\(=\left(3x+2-2x+7\right)^2=\left(x+9\right)^2\)

Thay x = - 19 vào A, ta có:

\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)

b, \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

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

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

\(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left(x+y\right)^2+6xy\)

\(=2\left(x+y\right)^3-6xy-3\left(x+y\right)^2+6xy\)

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

Thay x + y = 1 vào A, ta có:

\(A=2.1^3-3.1^2=-1\)

c, \(B=x^3+y^3+3xy\)

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

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

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

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

\(=\left(x+y\right)^3-3xy\left(x+y-1\right)\)

Thay x + y = 1 vào B, ta có:

\(B=1^3-3xy.\left(1-1\right)=1-3xy.0=1-0=1\)

d, \(C=8x^3-27y^3\)

\(=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)

\(=\left(2x-3y\right)\left(4x^2-12xy+9y^2+6xy\right)\)

\(=\left(2x-3y\right)\left[\left(2x-3y\right)^2+6xy\right]\)

\(=\left(2x-3y\right)^3+6xy\left(2x-3y\right)\)

Thay xy = 4 và 2x + 3y = 5 vào C, ta có:

\(C\)\(=5^3+6.4.5=125+120=245\)

Trả lời:

Bài 3:

\(A=x^2+x-2=\left(x^2+x+\frac{1}{4}\right)-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy GTNN của \(A=-\frac{9}{4}\Leftrightarrow x=-\frac{1}{2}\)

\(B=x^2+y^2+x-6y+2021\)

\(=x^2+y^2+x-6y+\frac{1}{4}+9+\frac{8047}{4}\)

\(=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-6y+9\right)+\frac{8047}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{8047}{4}\)\(\ge\frac{8047}{4}\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}}\)

Vậy GTNN của B = \(\frac{8047}{4}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

\(C=x^2+10y^2-6xy-10y+35\)

\(=x^2+9y^2+y^2-6xy-10y+25+10\)

\(=\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)+10\)

\(=\left(x-3y\right)^2+\left(y-5\right)^2+10\ge10\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-3y=0\\y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=15\\y=5\end{cases}}}\)

Vậy GTNN của C = 10 <=> \(\hept{\begin{cases}x=15\\y=5\end{cases}}\)

\(D=4x-x^2+5\)

\(=-\left(x^2-4x-5\right)\)

\(=-\left(x^2-4x+4-9\right)\)

\(=-\left[\left(x-2\right)^2-9\right]\)

\(=-\left(x-2\right)^2+9\le9\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2

Vậy GTLN của D = 9 <=> x = 2

\(E=-x^2-4y^2+2x-4y+3\)

\(=-x^2-4y^2+2x-4y-1-1+5\)

\(=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+5\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+5\le5\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}}\)

Vậy GTLN của D = 5  <=> \(\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}\)

Bài 3: 

a) \(\left(2-3x\right)^2-\left(3-x\right)^2=\left[\left(2-3x\right)-\left(3-x\right)\right]\left[\left(2-3x\right)+\left(3-x\right)\right]\)

\(=\left(-1-2x\right)\left(5-4x\right)\)

b) \(49\left(x-3\right)^2-9\left(x+2\right)^2\)

\(=\left[7\left(x-3\right)\right]^2-\left[3\left(x+2\right)\right]^2\)

\(=\left[\left(7x-21\right)-\left(3x+6\right)\right]\left[\left(7x-21\right)+\left(3x+6\right)\right]\)

\(=\left(4x-27\right)\left(10x-15\right)\)

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

d) \(2\left(x-3\right)+3\left(x^2-9\right)=2\left(x-3\right)+3\left(x-3\right)\left(x+3\right)\)

\(=\left(x-3\right)\left(3x+11\right)\)

e) \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)

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

f) \(1-2x+2yz+x^2-y^2-z^2=\left(x-1\right)^2-\left(y-z\right)^2\)

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

ta có : CK vuông góc DB (1)

AH vuông góc DB (2)

từ (1),(2) suy ra AH//CK (*)

xét tam giác vuông AHD và tam giác vuông CBK:ta có

góc H=góc K=90

góc ADH=góc CBK(slt)

suy ra 2 tam giác đó bằng nhau

suy ra AH=CK (*')

từ (*),(*') ta có tứ giác AHCK là hình bình hình

Xét tứ giác ABEC có 

AB//EC

AC//BE

Do đó: ABEC là hình bình hành

Suy ra: AC=BE

mà AC=BD

nên BE=BD

hay ΔBED cân tại B