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a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
a: \(9x^2-6x+3\)
\(=\left(9x^2-6x+1\right)+2\)
\(=\left(3x-1\right)^2+2\ge2\)
b: \(6x-x^2+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10\le10\)
\(A=\left(2n-1\right)^3-2n+1\)
\(A=8n^3-6n+6n-1-2n+1\)
\(A=8n^3-2n=2n\left(4n^2-1\right)\)
\(A=2n\left(2n+1\right)\left(2n-1\right)\)
\(A=\left(2n-1\right)2n\left(2n+1\right)⋮6\) ( 3 số tự nhiên liên tiếp)
Ta có:
P=x2+y2+z2+xy+yz+zx
\(\Rightarrow\) 2P= 2x2+2y2+2z2+2xy+2yz+2xz
= (x+y+z)2+x2+y2+z2
= 9+x2+y2+z2
Ta có x2+y2+z2\(\geq\) xy+yz+zx
\(\Leftrightarrow\) 3(x2+y2+z2)\(\geq\) x2+y2+z2+2xy+2yz+2zx
\(\Leftrightarrow\) 3(x2+y2+z2)\(\geq\) (x+y+z)2
\(\Leftrightarrow\) x2+y2+z2\(\geq\) \(\dfrac{\left(x+y+z\right)^2}{3}\) (1)
Từ đó suy ra: 9 + x2+y2+z2\(\geq\) 9+\(\dfrac{\left(x+y+z\right)^2}{3}\) = 9+\(\dfrac{9}{3}\)=9+3=12
\(\Rightarrow\) 2P\(\geq\)12
\(\Rightarrow\) P\(\geq\)6
Dấu = xảy ra khi x=y=z=1
Vậy MinP = 6 khi x=y=z=1
HELP ME!!!!!!!!!!!!!
THANKS TRƯỚC NHA!!!
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
Sửa đề: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Bài 1:
a) \(9x^2-6x+2\)
\(\Leftrightarrow9x^2-6x+1+1\)
\(\Leftrightarrow\left(3x-1\right)^2+1\)
Vì \(\left(3x-1\right)^2\ge0\forall x,1>0\)
\(\Rightarrow9x^2-6x+2\) luôn dương với mọi x.
b) \(x^2+x+1\)
\(\Leftrightarrow x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x,\dfrac{3}{4}>0\)
\(\Rightarrow x^2+x+1\) luôn dương với mọi x.
Bài 2 :
a) \(A=x^2-3x+5\)
\(\Leftrightarrow A=x^2-3x+2+3\)
\(\Leftrightarrow A=\left(x-2\right)\left(x-1\right)+3\)
Vì \(\left(x-2\right)\left(x-1\right)\ge0\forall x\) => \(A\ge3\)
Vậy GTNN A đạt được = 3 khi và chỉ khi x = 2 hoặc x = 1.
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(\Leftrightarrow B=4x^2-4x+1+x^2+4x+4\)
\(\Leftrightarrow B=5x^2+5\)
\(\Leftrightarrow B=5\cdot\left(x^2+1\right)\)
Vì \(x^2+1\ge1\forall x\)
=> GTNN của B đạt được = 5 khi và chỉ khi x = 0.
Bài 3 :
a) \(A=-x^2+2x+4\)
Làm tương tự ta có \(A_{MAX}=5\) khi và chỉ khi x = 1.
b) \(B=-x^2+4x\)
Làm tương tự ta có \(B_{MAX}=4\) khi và chỉ khi x = 2.
a)\(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1+y\right)\left(x-1-y\right)\)
b)\(x^3-4x^2-12x+27\)
\(=x^3-7x^2+9x+3x^2-21x+27\)
\(=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
c)\(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)