n *o biets

xy + 2x + y + 2 = y(x + 1) + 2(x + 1) = (x + 1).(y + 2)

x(x - 1) + x(x + 3) = x(x - 1 + x + 3) = x. ( 2x + 2) = 2x.(x + 1)

\(-4x^2+8x-4=-4\left(x^2-2x+1\right)=-4\left(x-1\right)^2\)

c: \(-4x^2+8x-4\)

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

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

a) \(xy+2x+y+2\)

\(=\left(xy+y\right)+\left(2x+2\right)\)

\(=y\left(x+1\right)+2\left(x+1\right)\)

\(=\left(y+2\right)\left(x+1\right)\)

b) \(x\left(x-1\right)+x\left(x+3\right)\)

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

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

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

c) \(-4x^2+8x-4\)

\(=-\left[\left(2x\right)^2-2.2x.2+2^2\right]\)

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

Đề là j 

Tính hay khai triển

đề là thực hiện phép toán ( 2 cách nha bạn

Trả lời:

Bài 2: 

a, \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)

Vậy ...

b, \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2000=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2000\end{cases}}\)

Vậy ...

c, \(2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

Vậy ...

d, \(\left(x+1\right)=\left(x+1\right)^2\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)

Vậy ...

Trả lời:

Bài 1: 

\(C=x-x^2=-\left(x^2-x\right)=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)

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

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

Vậy GTLN của C = 1/4 khi x = 1/2

\(E=4x^2+8x+y^2-4y+32=\left(2x\right)^2+8x+y^2-4y+4+4+24\)

\(=\left[\left(2x\right)^2+8x+4\right]+\left(y^2-4y+4\right)+24=\left(2x+2\right)^2+\left(y-2\right)^2+24\ge24\forall x\)

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

Vậy GTNN của E = 24 khi x = - 1; y = 2

Cảm ơn bạn nhiều :3

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

(4x - 1)(3 - x) - (2x + 3)(2x - 1) = 0

12x + \(4x^2\)- 3 + x - 4x + 2x - 6x + 3 = 0 

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

4x . 4x + 5x                                           = 0

  16x    + 5x                                           = 0

       21x                                                 = 0

           x                                                 = 0

2x(8x + 3) - (4x - 1)(4x + 1) = 13

16x + 6x   - 16x - 4x + 4x +1 = 13

6x +1                                     = 13

6x                                          = 12

  x                                           = 2

1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)

\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)

2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)

\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)

4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) 

\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)

\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)

e: Ta có: \(E=\left(x-8\right)^2+\left(x+7\right)^2\)

\(=x^2-16x+64+x^2+14x+49\)

\(=2x^2-2x+113\)

\(=2\left(x^2-x+\dfrac{113}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{225}{4}\right)\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)