1, Thay x = 1/3 ; y = -1/5 ta được 

\(=\dfrac{3.1}{9}-5\left(-\dfrac{1}{5}\right)+1=\dfrac{1}{3}+2=\dfrac{7}{3}\)

2, Thay x = -2 ; y = -1/2 ta được 

\(=5.4\left(-\dfrac{1}{2}\right)+3\left(-2\right)\left(-\dfrac{1}{2}\right)-\dfrac{2\left(-2\right).1}{4}\)

\(=-10+3+1=-6\)

E = 2x^2 - 5x -2 = 2( x^2 -5/2x -1) = 2(x^2 - 2.x.5/4 +25/16 - 41/16) = 2(x - 5/4 )^2 + 41/8

Vậy GTNN của biểu thức là 41/8 tại x = 5/4

F = x^2 + 5y^2 + 2xy -y +3 = (x^2 + 2xy +y^2) + (4y^2 - 2.2y.1/4 + 1/16) +47/16

(x + y)^2  + (2y - 1/4)^2 + 47/16 

Vậy GTNN của BT là 47/16 tại x = y = 1/8

a) -( x-y)² - (x-1)² -2 

GTLN = -2

a:

A = 2x - 2xy - 2x^2 - y^2

A = -(x^2 + 2xy + y^2) - (x^2 - 2x + 1) + 1

A = -(x + y)^2 - (x - 1)^2 + 1

Vì (x + y)^2 ≥ 0 ∀ x; y; (x - 1)^2 ≥ 0 ∀ x suy ra:

-(x + y)^2 ≤ 0 và (x - 1)^2 ≤ 0 ∀ x; y

A = -(x + y)^2 - (x - 1)^2 + 1 ≤ 1 ∀ x; y

Dấu bằng xảy ra khi x - 1 = 0 và x + y = 0

x - 1 = 0

x = 1

x + y = 0

y = - x

y = - 1

Vậy Amin = 1 khi x = 1; y = - 1

A = (x^2 + 2x + y^2

5A:

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

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

=(y-2x+1)(y+2x-1)

b: \(x^2-y^2-5y+5x\)

=(x-y)(x+y)+5(x-y)

=(x-y)(x+y+5)

6A:

a: \(x^2-8x+7\)

\(=x^2-x-7x+7\)

=x(x-1)-7(x-1)

=(x-1)(x-7)

b: \(2x^2-5x+2\)

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

=2x(x-2)-(x-2)

=(x-2)(2x-1)

c: \(x^3-5x^2+8x-4\)

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

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

\(a,A=5x+8xy+5y=(5x+5y)+8xy\)

\(=5(x+y)+8xy\)

\(=5\cdot\frac{2}{5}+8\cdot(-1)=2+(-8)=-6\)

\(b,B=2xy+7xyz-2xz\)

\(=2\cdot\frac{3}{7}y+7\cdot\frac{3}{7}yz-2\cdot\frac{3}{7}z\)

\(=\frac{6}{7}y+3yz-\frac{6}{7}z\)

\(=\frac{6}{7}y+3\cdot(-1)-\frac{6}{7}z\)

\(=\frac{6}{7}y+(-3)-\frac{6}{7}z\)

Làm nốt :v

a)

A=\(5\left(x+y\right)+8xy\) 

 \(=5.\frac{2}{5}+8.\left(-1\right)\) 

  \(=2-8\) 

   \(=-6\) 

Vậy.......

hc tốt

có làm mới có ăn nha em

a, \(A=x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi x=-1/2

Vậy Amin=3/4 khi x=-1/2

b,\(B=2x^2-5x-2\)

\(\Rightarrow2B=4x^2-10x-4=\left(4x^2-10x+\frac{25}{4}\right)-\frac{41}{4}=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\)

Vì \(\left(2x-\frac{5}{2}\right)^2\ge0\Rightarrow2B=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\ge-\frac{41}{4}\Rightarrow B\ge-\frac{41}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmin=-41/8 khi x=5/4

c,\(C=x^2+5y^2+2xy-y+3=\left(x^2+2xy+y^2\right)+\left(4y^2-y+\frac{1}{16}\right)+\frac{47}{16}=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\)

Vì\(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(2y-\frac{1}{4}\right)^2\ge0\end{cases}}\Rightarrow\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2\ge0\)

\(\Rightarrow C=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\ge\frac{47}{16}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\2y-\frac{1}{4}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{8}\\y=\frac{1}{8}\end{cases}}}\)

Vậy Cmin=47/16 khi x=-1/8,y=1/8