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a, \(A=5x-x^2=-x^2+5x=-x^2+2x\cdot2,5-\dfrac{25}{4}+\dfrac{25}{4}\)
\(=-\left(x-2,5\right)^2+\dfrac{25}{4}\)
Có: \(-\left(x-2,5\right)^2\le0\forall x\)
=> \(-\left(x-2,5\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
''='' xảy ra khi \(x-2,5=0\Rightarrow x=2,5\)
Vậy \(A_{MAX}=\dfrac{25}{4}\Leftrightarrow x=2,5\)
b, \(B=x-x^2=x^2-x=x^2-2\cdot x\cdot\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Lập luận như câu a
c, \(C=4x-x^2+3=-x^2+2\cdot x\cdot2-4+7\)
\(=-\left(x-2\right)^2+7\)
Vì \(-\left(x-2\right)^2\le0\forall x\)
=> \(-\left(x-2\right)^2+7\le7\)
Dấu ''='' xảy ra khi và chỉ khi x = 2
Vậy \(C_{MAX}=7\Leftrightarrow x=2\)
d, \(D=-x^2+6x-11=-x^2+2\cdot x\cdot3-9-2\)
\(=-\left(x-3\right)^2-2\)
Vì \(-\left(x-3\right)^2\le0\forall x\)
=> \(-\left(x-3\right)^2-2\le-2\)
Dấu ''='' xảy ra khi và chỉ khi x - 3 = 0 => x = 3
Vậy \(D_{MAX}=-2\Leftrightarrow x=3\)
e, \(E=5-8x-x^2=-x^2-8x+5=-x^2-2\cdot x\cdot4-16+21\)
\(=-\left(x+4\right)^2+21\)
Lập luận như trên
f, \(F=4x-x^2+1=-x^2+4x+1=-x^2+2\cdot x\cdot2-4+5\)
\(=-\left(x-2\right)^2+5\)
Tượng tự mấy ý trc
Bài 2:
a: \(A=-3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=2/3
b: \(B=-x^2+5x+3\)
\(=-\left(x^2-5x-3\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)
Dấu '=' xảy ra khi x=5/2
c) Đặt \(t=x^2+x+1\) thì
\(t\left(t+1\right)-12=t^2+t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
d) \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+11\) thì
\(\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Rồi nha bạn 
phân tích đa thức thành nhân tử
a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)
\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) \(x^2+2xy+y^2-x-y-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y-4\right)\left(x+y+3\right)=0\)
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) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
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) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
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) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(a,A=-1+3-5+7-9+...-2013+2015-2017=\left(-1+3\right)+\left(-5+7\right)+...+\left(-2013+2015\right)-2017\)\(=2+2+..+2-2017\)
\(=2.504-2017=-1009\)
\(b,B=2-4+6-8+...+2014-2016+2018\)\(=2+\left(-4+6\right)+\left(-8+10\right)+...+\left(-2016+2018\right)==2+2+...+2\)\(=2+503.2=1008\)
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, a.(b+c) +3b+3c = a.(b+c) +3.(b+c)
= (a+3)(b+c)
b, a(m-n)+m-n = (a+1)(m-n)