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Từ I vẽ đường thẳng II' // BC
Từ D vẽ đường thẳng DD' // BC
=> II' // DD' . Mà I là trung điểm của DE
=> EI' = I'D' ( 1 )
Vì \(\Delta\)ABC cân tại A có DD' // BC => DB = D'C ( 2 )
Mà AD = CE => AE = DB ( 3 )
Từ ( 2 ) và ( 3 ) => D'C = AE ( 4 )
Từ ( 1 ) và ( 4 ) => AI' = 'IC
\(\Delta\)AKC có II' // KC ; AI' = I'C
=>AI = IK ( Đpcm )
a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)
Vậy Max(A) = 7 khi x = -1
b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy Max(B) = 11 khi x = 3
c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{1}{3}\right)^2=0\Rightarrow x=\frac{1}{3}\)
Vậy Max(C) = 1/3 khi x = 1/3
d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy Max(D) = 17/4 khi x = 3/2
e) \(E=3-2x^2+2xy-y^2-2x\)
\(E=-\left(x^2-2xy+y^2\right)-\left(x^2+2x+1\right)+4\)
\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)
Vậy Max(E) = 4 khi x = y = -1
a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x+2\right)^2+\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x+2\right)\left(x+2+x-2\right)+\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x+2\right)+2x\left(x-2\right)\)
\(=2x\left(x+2+x-2\right)\)
\(=2x\cdot2x=4x^2\)
b) \(2x^2-2xy-4y^2\)
\(=\left(2x^2-4xy\right)+\left(2xy-4y^2\right)\)
\(=2x\left(x-2y\right)+2y\left(x-2y\right)\)
\(=\left(2x+2y\right)\left(x-2y\right)\)
\(=2\left(x+y\right)\left(x-2y\right)\)
c) \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
d) \(4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
\(=4\left(x-2y\right)\left(x-2y\right)\)
\(=4\left(x-2y\right)^2\)
x2 + 2y2 + z2 - 2xy - 2y - 4z + 5 = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + ( z2 - 4z + 4 ) = 0
<=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2 = 0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )
Thay ( 1 ) vào A , ta được :
\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)
Vậy A = 2
Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)
Ta có: \(x^2+x-6=\left(x-2\right)\left(x+3\right)\)
Đặt \(A\left(x\right)=x^3+ax^2-bx+12\)
Để A(x) chia hết cho \(x^2+x-6\) thì mọi nghiệm của \(x^2+x-6\) đều là nghiệm của A(x)
=> x = 2 và x = -3 là 2 nghiệm của A(x)
Ta có: \(\hept{\begin{cases}A\left(2\right)=2^3+4a-2b+12=0\\A\left(-3\right)=\left(-3\right)^3+\left(-3\right)^2a-\left(-3\right)b+12=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}4a-2b=-20\\9a+3b=15\end{cases}}\Leftrightarrow\hept{\begin{cases}2a-b=-10\\3a+b=5\end{cases}}\)
\(\Rightarrow2a-b+3a+b=-10+5\)
\(\Leftrightarrow5a=-5\Rightarrow a=-1\Rightarrow b=8\)
Vậy a = -1 ; b = 8
1) \(x^5-x^4-1\)
\(=x^5-x^4-1+x^3-x^3+x^2-x^2+x-x\)
\(=\left(x^5-x^3-x^2\right)-\left(x^4-x^2-x\right)+\left(x^3-x-1\right)\)
\(=x^2\left(x^3-x-1\right)-x\left(x^3-x-1\right)+\left(x^3-x-1\right)\)
\(=\left(x^3-x-1\right)\left(x^2-x+1\right)\)
2) \(x^8-3x^4+1\)
\(=x^8-3x^4+1+x^4-x^4\)
\(=\left(x^8-2x^4+1\right)-x^4\)
\(=\left(x^4-1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2-1\right)\left(x^4+x^2-1\right)\)
a. x4 - 27x = x ( x3 - 33 ) = = x ( x - 3 ) ( x2 + 3x + 32 ) = x ( x - 3 ) ( x2 + 3x + 9 )
b. x3 + 2x2 + 2x + 1 = ( x3 + 13 ) + ( 2x2 + 2x ) = ( x + 1 ) ( x2 - x + 1 ) + 2x ( x + 1 ) = ( x + 1 ) ( x2 + x + 1 )
c. 4x - 4y + x2 - 2xy + y2 = 4 ( x - y ) + ( x - y )2 = ( x - y ) ( x - y + 4 )
a
\(x^4-27x\)
\(=x\left(x^3-27\right)\)
\(=x\left(x^3-3^3\right)\)
\(=x\left(x-3\right)\left(x^2+3x+9\right)\)
b
\(x^3+2x^2+2x+1\)
\(=x^3+x^2+x^2+x+x+1\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+1\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
c
\(4x-4y+x^2-2xy+y^2\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(x-y+4\right)\)
A = -5 - (x - 1)(x + 2)
= -5 - [x(x + 2) - 1(x + 2)]
= -5 - (x2 + 2x - x - 2)
= -5 - x2 - 2x - x + 2 = -5 - x2 - x + 2 = (-5 + 2) - x2 - x = -3 - x2 - x
= -(x + x2 + 3) = -(x2 + x + 3)
= -[x2 + 2.x.1/2 + (1/2)2 ] - 11/4
= -(x + 1/2)2 - 11/4
Vì (x + 1/2)2 \(\ge\)0\(\forall\)x
=> -(x + 1/2)2 \(\le\)0\(\forall\)x
=> \(-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}\ge-\frac{11}{4}\forall x\)
Dấu " = " xảy ra khi (x + 1/2)2 = 0 => x = -1/2
Vậy Amax = -11/4 khi x = -1/2
\(A=-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-\left(x^2+x-2\right)\)
\(=-5-x^2-x+2\)
\(=-x^2-x-3\)
\(=-x^2-x-\frac{1}{4}-3+\frac{1}{4}\)
\(=-\left(x^2+x+\frac{1}{4}\right)-\frac{11}{4}\)
\(=-\left(x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right)-\frac{11}{4}\)
\(=-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}\)
Ta có \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\\(-\left(x+\frac{1}{2}\right)^2\le0\)
\(-\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)
Dấu = xảy ra
\(\Leftrightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
