sau 3/11h thì kim phút đuổi kịp kim h

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)(1)

Đặt \(x^2+5x=t\)

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

\(=t^2+3t-t-3=t\left(t+3\right)-\left(t+3\right)\)

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

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

hay \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)\(=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)

3x - 6y = 3( x - 2y)

=3.(x-2y)

\(x+y+z=0\)

\(\Leftrightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy.\left(-z\right)\)

\(\Leftrightarrow x^3+y^3+z^3=3xyz\left(đpcm\right)\)

Ta có \(x+y+z=0\Leftrightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-z^3\)

\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)=3xyz\left(đpcm\right)\)

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

\(=-\left(x^2+8x+16-17\right)\)

\(=-\left[\left(x+4\right)^2-17\right]\)

\(=-\left(x+4\right)^2+17\le17\)

Vậy \(A_{max}=17\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(B=-3x^2+6x-5\)

\(=-3\left(x^2-2x+\frac{5}{3}\right)\)

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

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

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

Vậy \(B_{mã}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)

A= {(x^2-8x-(2^2)+5

  =(x-2)^2+5

  =(x-2)^2+5>/0+5=5

  Dấu = xảy ra khi (x-2)^2=0

                  suy ra x=2

B=-3(x^2-2x+1)-6

  =-3(x+1)^2-6>/6

  gtln là 6 khi x= 1                   suy ra x=

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

\(=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)\)

\(=2\left(x^2+\frac{3}{2}x+\frac{9}{16}-\frac{1}{16}\right)\)

\(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge\frac{-1}{8}\)

Vậy \(A_{min}=\frac{-1}{8}\Leftrightarrow x+\frac{3}{4}=0\Leftrightarrow x=-\frac{3}{4}\)

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

b) \(x^2+16x+64=\left(x+8\right)^2\)

c) \(x^3-8y^3=x^3-\left(2y\right)^3\)

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

d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)

\(D=8x^2-2x-1\)

\(=8\left(x^2-\frac{1}{4}x-\frac{1}{8}\right)\)

\(=8\left(x^2-\frac{1}{4}x+\frac{1}{64}-\frac{9}{64}\right)\)

\(=8\left[\left(x-\frac{1}{8}\right)^2-\frac{9}{64}\right]\)

\(=8\left(x-\frac{1}{8}\right)^2-\frac{9}{8}\ge\frac{-9}{8}\forall x\left(đpcm\right)\)

D=\(8x^2-2x+1\)

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

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

Nhóm lại đc 1 cái mũ 2 trừ bn đó

chia 2 ra đc >= -9/8 là ĐPCM