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a) ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014
Đăngt thức xay ra khi x=y=1
\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)
=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)
= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+5}\)
= \(3+\dfrac{2}{x^2-2x+1+4}\)
= \(3+\dfrac{2}{\left(x-1\right)^2+4}\)
vì (x-1)2 ≥ 0 ∀ x
⇔ (x-1)2 +4 ≥ 4
⇔\(\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{1}{2}\)
⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)
⇔ A \(\le\dfrac{7}{2}\)
⇔ Min A =\(\dfrac{7}{2}\)
khi x-1=0
⇔ x=1
vậy ....
Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)
\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(B=2-\dfrac{3}{x^2-8x+16+6}\)
\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)
\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)
a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi x=3/2
c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)
=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)
=>C>=-12/7
Dấu '=' xảy ra khi x=1/2
c)\(C=2x^2+y^2+2xy-8x+2019\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+2003\)
\(=\left(x+y\right)^2+\left(x-4\right)^2+2003\)
Vì \(\left(x+y\right)^2\ge0,\forall x,y\)
\(\left(x-4\right)^2\ge0,\forall x\)
Nên \(\left(x+y\right)^2+\left(x-4\right)^2\ge0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-4\right)^2+2003\ge2003\)
\(\Rightarrow min_C=2003\) khi \(\left\{{}\begin{matrix}x-4=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)
d)\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(t=x^2+5x\) ta được:
\(D=\left(t-6\right)\left(t+6\right)=t^2-36\)
Vì \(t^2\ge0,\forall t\)
\(\Rightarrow t^2-36\ge-36\)
\(\Rightarrow min_D=-36\) khi \(t=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy Min D = -36 khi \(x\in\left\{0;5\right\}\)
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) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)