Chứng minh rằng với mọi x thuộc Q thì giá trị của đa thức:
M=(x+2)(x+4)(x+6)(x+8)+16 là bình phương của một số hữu tỉ.
M = (x + 2)(x + 4)(x + 6)(x + 8) + 16
M = [(x + 2)(x + 8)][(x + 4)(x + 6)] + 16
M = (x^2 + 2x + 8x + 16)(x^2 + 4x + 6x + 24) + 16
M = (x^2 + 10x + 16)(x^2 + 10x + 24) + 16
Đặt t = x^2 + 10x + 20
M = (t - 4)(t + 4) + 16
M = t^2 - 16 + 16 = t^2
Vậy ta có đpcm
\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)\left(x-3\right)\left(x+3\right)=\left(x-3\right)^2\left(x+3\right)\)
\(\left(3x-2\right)\left(3x+2\right)-\left(4-3x\right)^2=9x^2-4-16+24x-9x^2=24x-20=4\left(6x-5\right)\)
\(\left(2x+5\right)^2-\left(1+2x\right)\left(2x-1\right)=-3\)
\(4x^2+20x+25-4x^2+1=-3\)
\(20x=-3-25-1\)
\(20x=-29\)
\(x=-\frac{29}{20}\)
a) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
b) (a + b - c)2 = a2 + b2 + c2 + 2ab - 2ac - 2bc
c) (a - b - c)2 = a2 + b2 + c2 - 2ab - 2ac + 2bc
a)
( a + b + c)2 = [(a + b) + c ]2
= (a + b)2 + 2(a + b)c + c2
= a2 + 2ab + b2 + 2ac +2bc + c2
= a2 + b2 + c2 + 2ab + 2ac + 2bc
b)
(a + b - c)2 = [(a + b) - c ]2
= (a + b)2 - 2(a + b)c + c2
= a2 + 2ab + b2 - 2ac - 2bc + c2
= a2 + b2 + c2 + 2ab - 2ac - 2bc
c)
(a - b - c)2 = [(a - b) - c ]2
= (a - b)2 - 2(a - b)c + c2
= a2 - 2ab + b2 - 2ac + 2bc + c2
= a2 + b2 +c2 - 2ab - 2ac + 2bc
\(x^6+3x^2y^2+y^6=\left(x^6+y^6\right)+3x^2y^2=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)+3x^2y^2\)
\(=x^4-x^2y^2+y^4+3x^2y^2=x^4+2x^2y^2+y^4=\left(x^2+y^2\right)^2=1^2=1\)
\(P=x^2-2xy+6y^2-12x+3y+45\)
\(=x^2-2x\left(y+6\right)+\left(y+6\right)^2-\left(y+6\right)^2+6y^2+3y+45\)
\(=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+\left(5y^2-9y+9\right)\)
\(=\left(x-y-6\right)^2+5\left(y-\frac{9}{10}\right)^2+\frac{99}{20}\)
\(\ge\frac{99}{20}\) . Đẳng thức xảy ra khi y = 9/10, x = 69/10
Vậy min P = 99/20 tại x = 69/10, y = 9/10
\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(\Rightarrow\left[\left(a+d\right)+\left(b+c\right)\right]\left[\left(a+d\right)-\left(b+c\right)\right]-\left[\left(a-d\right)-\left(b-c\right)\right]\left[\left(a-d\right)+\left(b-c\right)\right]=0\)
\(\Rightarrow\left(a+d\right)^2-\left(b+c\right)^2-\left(a-d\right)^2+\left(b-c\right)^2=0\)
\(\Rightarrow a^2+d^2+2ad-b^2-c^2-2bc-a^2-d^2+2ad+b^2+c^2-2bc\)
\(\Rightarrow4ad-4bc\)
\(\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(t=x^2+x\) ta có pt trở thành:
\(t^2+4t=12\Leftrightarrow t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)\(\Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)+4=8\)
\(\Leftrightarrow\left(x^2+x+2\right)^2=8\)
\(\left[{}\begin{matrix}x^2+x+2=8\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\x^2+x+2=-8\left(loai\right)\end{matrix}\right.\)
\(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)
\(\Leftrightarrow30x=60\)
\(\Leftrightarrow x=2\)
A = 1002 - 992 + 982 - 972 + . . . + 22 - 12
= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)
= 199 + 195 + . . . + 3
= 5050
B = 3(22 + 1)(24 + 1) . . . (264 + 1) + 1
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1)(264 + 1) + 1
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (28 - 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (216 - 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (232 - 1)(232 + 1)(264 + 1) + 1
= (264 - 1)(264 + 1) + 1
= 2128 - 1 + 1
= 2128