làm nhan hộ với 8 h mink phải nộp rồi

\(ax^2+bx^2-bx-ax+cx^2-cx\)

\(=\left(ax^2-ax\right)+\left(bx^2-bx\right)+\left(cx^2-cx\right)\)

\(=ax\left(x-1\right)+bx\left(x-1\right)+cx\left(x-1\right)\)

\(=\left(x-1\right)\left(ax+bx+cx\right)\)

\(=x\left(x-1\right)\left(a+b+c\right)\)

ax²+bx²-bx-ax+cx²-cx

= (ax² - ax ) + (bx² -bx ) + ( cx² - cx )

= a(x) + b(x) + c(x)

= (x)(a+b+c)

a⁴ + 4 = (a²)² + 4 = (a²)² + 4a² + 4 - 4a²

                             = (a² + 2)² - (2a)²

                             = (a² - 2a + 2)(a² + 2a + 2)

a⁴+4=

a⁴+ 4a²+4-4a²=(a²+2)²-4a²=(a²+2+2x).(a²+2-2x)

khó vaiz

thế mới hỏi

Ta có A = x² - 6x + 10

= x² - 6x + 9 + 1 

= (x - 3)² + 1 \(\ge\)1 

Dấu "=" xảy ra <=>x - 3 = 0 => x = 3

Vậy Min A = 1 <=> x = 3

\(A=2x^3+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)

\(=2x^3+3\left(x^2-1\right)-5x^2-5x\)

\(=2x^3+3x^2-3-5x^2-5x\)

\(=2x^3-2x^2-5x-3\)

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

\(=125-150x+60x^2-8x^3-\left(25-9x^2\right)\)

\(=125-150x+60x^2-8x^3-25+9x^2\)

\(=100-15x+69x^2-8x^3\)

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

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

\(=\left(x+2\right).5x\)

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

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

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

\(E=\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-8+x\left(x^2-1\right)+6x^2-18x\)

\(=x^3-6x^2+12x-8+x^3-x+6x^2-18x\)

\(=-7x-8\)

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

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

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

\(=-3\)

a) Ta có A = x² - 2x - 1 = (x² - 2x + 1) - 2 = (x - 1)² - 2 \(\ge\) -2 

Dấu "=" xảy ra <=> x - 1 = 0 => x = 1

Vậy Min A = -2 <=> x = 1 

b) Ta có B = 4x² + 4x + 8 = (4x² + 4x + 1) + 7 = (2x + 1)² + 7 \(\ge\)7

Dấu |"=" xảy ra <=> 2x + 1 = 0 => x = -1/2

Vậy Min B = 7 <=> x = -1/2

c) Ta có C = 3x - x² + 2

                 = -(x² - 3x - 2)

                = -(x² - 3x + 9/4 - 9/4 - 2)

                = -[(x - 3/2)² - 17/4)

                 = -(x - 3/2)² + 17/4 \(\le\frac{17}{4}\)

Dấu "=" xảy ra <=> x - 3/2 = 0 => x = 3/2

Vậy Max C = 17/4 <=> x = 3/2

d) Ta có D = -x² - 5x = -(x² + 5x) = -(x² + 5x + 25/4 - 25/4) = -(x + 5/2)² + 25/4 \(\ge\frac{25}{4}\)

Dấu "=" xảy ra <=> x + 5/2 = 0 => x = -5/2

Vậy Max D = 25/4 <=> x = -5/2

e) Ta có E = x² - 4xy + 5y² + 10x - 22y + 28

                  = (x² - 4xy + 4y²) + 10x - 20y + y² - 2y + 28

                 = (x - 2y)² + 10(x - 2y) + 25 + (y² - 2y + 1) + 2

                 = (x - 2y + 5) + (y - 1)² + 2 \(\ge\)2

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min E = 2 <=> x = -3 ; y = 1

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

Dấu \(=\)xảy ra khi \(x=1\). Vậy GTNN của \(A\)là \(-2\).

\(B=4x^2+4x+8=4x^2+4x+1+7=\left(2x+1\right)^2+7\ge7\)

Dấu \(=\)xảy ra khi \(x=\frac{-1}{2}\). Vậy GTNN của \(B\)là \(7\).

\(C=-x^2+3x+2=-x^2+2.\frac{3}{2}x-\left(\frac{3}{2}\right)^2+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\)

Dấu \(=\) xảy ra khi \(x=\frac{3}{2}\). Vậy GTLN của \(C\)là \(\frac{17}{4}\).

\(D=-x^2-5x=-x^2-2.\frac{5}{2}x-\left(\frac{5}{2}\right)^2+\frac{25}{4}=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu \(=\)xảy ra khi \(x=\frac{-5}{2}\). Vậy GTLN của \(D\) là \(\frac{25}{4}\).

\(E=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2+4y^2+25-4xy+10x-20y+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\). Vậy GTNN của \(E\) là \(2\).