\(A=x^2+2y^2-2xy+4x-2y+2017\)

\(A=x^2+y^2+y^2-2xy+4x-4y+2y+2017\)

\(A=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+\left(y^2+2y+1\right)+2012\)

\(A=\left(x-y\right)^2+2\left(x-y\right)2+2^2+\left(y+1\right)^2+2012\)

\(A=\left(x-y+2\right)^2+\left(y+1\right)^2+2012\)

Vì \(\left(x-y+2\right)^2\ge0\forall x;y,\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow A\ge2012\forall x;y\)

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

Vậy A_min = 2012 khi và chỉ khi x = -3; y = -1

\(3x\left(x+2\right)-2\left(x^2-5x\right)-x\left(x+10\right)=12\)

\(\Rightarrow3x^2+6x-2x^2+10x-x^2-10x=12\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

Vậy x = 2

a) x^2.(x^2+4) - x^2 + 4

= x^4 + 4.x^2 - x^2 + 4

= x^4 + 3.x^2 + 4

= x^4 + 4.x^2 + 4 - x^2

= (x^2+2)^2 - x^2

= (x^2+2+x).(x^2+2-x)

b) x^2.(x+4)^2 - (x+4)^2 - (x^2-1)

= (x+4)^2.(x^2-1) - (x^2-1)

= (x^2-1).[(x+4)^2- 1]

= (x+1).(x-1).(x+3).(x+5)

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

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

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

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

\(b,x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

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

\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

Bài giải
a, + I là trung điểm BC nên BI=IC=BC2=2a:2=a=AB=CDBI=IC=BC2=2a:2=a=AB=CD

+ CM: △ABI=△DCI△ABI=△DCI (cgc)

~~> AI=DIAI=DI (2 cạnh tương ứng) ~~> △IAD△IAD cân ở I ~~> A1ˆ=D1ˆA1^=D1^ (1)

+ △IAD△IAD có Hk là đường trung bình nên HK // AD (2)

+ Từ (1) và (2) ta có AHKDAHKD là hình thang cân 

b, + △ABI△ABI vuông ở B theo pytago có BI2+BA2=AI2BI2+BA2=AI2. Hay AI2=2a2⟹AI=2a2−−−√=DIAI2=2a2⟹AI=2a2=DI (theo phần a AI=DI)

+ H là trung điểm AI nên : AH=AI2=2a2−−−√2AH=AI2=2a22 

+ Tương tự có KD=2a2−−−√2KD=2a22

+ Ta có AD=BC=2aAD=BC=2a

+ HK là đường trung bình△IAD△IAD nên HK=AD2=aHK=AD2=a

+ Chu vi hình thang HKDA là KD+DA+AH+HD=2a2−−−√2+2a2−−−√2+a+2a=2a2−−−√+3a

a) 5 (x-2)-3x=0

=>5x-10-3x=0

=>2x-10=0

=>x=5

b) => x^2=49 =>x=+-7

c)3x^2 +7x-18=0

=> ( vô ngiệm)

d)3x (x^2-16)=0

=>3x=0 hoặc x^2-16=0

=>x=0 hoặc x=+-4

K mk nha bn, 

1) 5(x-2)-3x=0

=> 5x-10-3x=0

=> 2x-10=0

=> 2x=10

=>x=5

2) x²-49=0

=> x²=49

=> x=+-7

3) 3x²+7x-18=0

=> \(x\in\varnothing\)

4) 3x(x²-16)=0

=> 3x(x-4)(x+4)=0

=>\(\hept{\begin{cases}x=0\\x-4=0\\x+4=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)

vậy x=0 hoặc x=+-4

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)=3\left[\left(x^4-x^3+x^2\right)+\left(x^3+1\right)\right]-\left(x^2+x+1\right)\)

\(=3\left[x^2\left(x^2-x+1\right)+\left(x+1\right)\left(x^2-x+1\right)\right]-\left(x^2+x+1\right)\)

\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(3x^2-3x+3-1\right)\)

\(=\left(x^2+x+1\right)\left(3x^2-3x+2\right)\)