Giups mk vs ạ ai nhanh mk tick nha

Lời giải:
\(x^2+3y^2+10x-14y-2xy=11\)

$\Leftrightarrow (x^2-2xy+y^2)+2y^2+10x-14y=11$

$\Leftrightarrow (x-y)^2+10(x-y)+25+(2y^2-4y+2)=38$

$\Leftrightarrow (x-y+5)^2+2(y-1)^2=38$

$\Rightarrow (x-y+5)^2=38-2(y-1)^2\leq 38$

$\Rightarrow -\sqrt{38}\leq x-y+5\leq \sqrt{38}$

$\Leftrightarrow -\sqrt{38}-5\leq x-y\leq \sqrt{38}-5$
Vậy $A_{\min}=-\sqrt{38}-5$ và $A_{\max}=\sqrt{38}-5$

\(1,Sửa:A=4x^4+4x^2y+y^2+2=\left(2x^2+y\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow2x^2+y=0\Leftrightarrow x^2=-\dfrac{y}{2}\\ 2,B=\left(x+y\right)^2+\left(y+1\right)^2+12\ge12\\ B_{min}=12\Leftrightarrow\left\{{}\begin{matrix}x=-y=1\\y=-1\end{matrix}\right.\)

\(\left|x\right|=2\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Khi x = 2:

\(E=5.2^2-3.2.\left(-1\right)\\ E=20-\left(-6\right)=26\)

Khi x = -2

\(E=5.\left(-2\right)^2-3.\left(-2\right).\left(-1\right)\\ E=20-6=14\)

x|=2⇒[x=2x=−2

Trg khi x = 2:

Trong khi x = -2

E=x^2+2y^2-2xy+2x-10y

=x²+y²-2xy+y²-8y+16+2x-2y-16

=(x-y)²+(y-4)²+2.(x-y)-16

=(x-y)²+2(x-y)+1+(y-4)²-17

=(x-y+1)²+(y-4)²-17 \(\ge\)-17

Dấu "=" xảy ra khi: y=4; x=3

Vậy GTNN của E là -17 tại x=3;y=4

\(E=x^2+2y^2-2xy+2x-10y\)

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

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

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

\(=\left[x-\left(y-1\right)\right]^2+2y^2-10y-y^2+2y-1\)

\(=\left(x-y+1\right)^2+y^2-8y-1=\left(x-y+1\right)^2+\left(y^2-2.y.4+16\right)-17\)

\(=\left(x-y+1\right)^2+\left(y-4\right)^2-17\)

Vì \(\left(x-y+1\right)^2\ge0;\left(y-4\right)^2\ge0=>E=\left(x-y+1\right)^2+\left(y-4\right)^2-17\ge-17\) (với mọi x;y)

Dấu "=" xảy ra \(< =>\hept{\begin{cases}\left(x-y+1\right)^2=0\\\left(y-4\right)^2=0\end{cases}< =>\hept{\begin{cases}x-y=-1\\y=4\end{cases}}< =>x=3;y=4}\)

Vậy minE=-17 khi x=3;y=4

a) =(5x)^2-2*5x+1+3

   =(5x-1)^2+3

suy ra min=3

b) = -(x^2-2x+1)-1

    =-(x^2-1)^2-1

suy ra Max=-1

c)=(x^2-2x+1)+(y^2-4y+4)+1

  =(x^2-1)^2+(y^2-2)^2+1

suy ra Min=1

# mk ko chắc lắm đâu

B=(x^2-6x+9)-8

B=(x-3)^2-8

Vì (x-3)^2\(\ge0\forall x\)

-> (x-3)-8\(\ge-8\forall x\)

Dấu = xảy ra<=> x-3=0<=>x=3

C=2x^2-10x+1

C=2(x^2-5x+6,25)-11,5

C= 2(x-2,5)^2-11,5

Vì 2(x-2,5)^2\(\ge0\forall x\)

->2(x-2,5)^2-11,5\(\ge-11,5\forall x\)

Dấu = xẩy ra<=> x-2,5=0<=>x=2,5

Vậy Min C là -11,5 <=> x=2,5

D= x^2+10-25

D=(x^2+10+25)-50

D=(x+5)^2-50

Vì (x-5)^2 \(\ge0\forall x\)

-> (x-5)^2-50\(\ge-50\forall x\)

Dấu = xẩy ra <=> x-5=0<=>x=5

Vậy Min D là -50 <=>x=5

Tìm Max

B= 5x-x^2

B=-(x^2-5x+25/4)-25/4

B= -(x-5/2)^2-25/4

Vì -(x-5/2)^2\(\le0\forall x\)

-> -(x-5/2)^2-25/4\(\le\)-25/4

Dấu = xẩy ra <=> x-5/2=0<=>x=5/2

Vậy Max B là -25/4 <=> x=5/2

C=-x^2-6x+10

C=-(x^2+6x+9)+19

C= -(x+3)^2+19

Vì -(x+3)^2\(\le\)0

=> -(x+3)^2+19\(\le\)19

Dấu = xảy ra <=> x+3=0<=>x=-3

D= -2x^x+8x+12

D=-2(x^2-4x+4)+20

D=-2(x-2)^2 +20

 Vì -2(x-2)^2\(\le\)0

=> -2(x-2)^2+20\(\le\)20

Dấu= xẩy ra<=> x-2=0<=>x=2

Vậy Max D là 20<=>x-2

C = x¹⁴ - 10x¹³ + 10x¹³ -10x¹¹ + ... + 10x¹² -10x + 10 

    = x¹⁴ - ( x + 1 )x¹³ + ( x + 1)x¹² -... - ( x + 1)x + 10 + 1

    =x¹⁴ -x¹⁴ - x¹³ + x¹³ + x¹² - ...- x² - x + 10 + 1 

     = 1

Không chắc lắm

Câu 2: 

a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)

\(=3x-\left|x-5\right|\)

\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)

b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)