Phân thức đại số - Hỏi đáp

Đặt \(Q=\dfrac{2011}{2a^2+2b^2+2008}\)

Ta có:

\(\dfrac{a+b}{2}=1=>a+b=2=>a=2-b\)

Thay a=2-b vào Q ta được:

\(Q=\dfrac{2011}{2a^2+2\left(2-a\right)^2+2008}\)

=\(\dfrac{2011}{2a^2+2\left(4-4a+a^2\right)+2008}\)

=\(\dfrac{2011}{2a^2+8-8a+2a^2+2008}\)

=\(\dfrac{2011}{4a^2-8a+2016}\)

=\(\dfrac{2011}{4a^2-8a+4+2012}\)

=\(\dfrac{2011}{4\left(a^2-2a+1\right)+2012}\)

=\(\dfrac{2011}{4\left(a-1\right)^2+2012}\)

Vì \(2a^2+2b^2+2008>0với\forall a,b\)

nên để Q đạt GTLN thì \(2a^2+2b^2+2008\)đạt GTNN hay \(4\left(a-1\right)^2+2012\)đạt GTNN

Mặt khác \(4\left(a-1\right)^2\)\(\ge\)0 với \(\forall\)a

Do đó\(4\left(a-1\right)^2+2012\) \(\ge\)0 với \(\forall\)a

Dấu "=" xảy ra <=> a-1=0<=>a=1

Mà a+b=2=>b=1

Vậy GTN của \(Q=\dfrac{2011}{2a^2+2b^2+2008}\)là \(\dfrac{2011}{2012}\)khi a=b=1

Cách 2 :

Ta có : \(\dfrac{a+b}{2}=1\Rightarrow a+b=2\)

Mặt khác : Với \(\forall a,b\) thì : \(a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2=4\)

\(\Rightarrow2a^2+2b^2+2008\ge2012\)

\(\Rightarrow\dfrac{2011}{2a^2+2b^2+2008}\le\dfrac{2011}{2012}\)

Dấu '' = '' xảy ra khi a = b = 1

Vậy GTLN của biểu thức là \(\dfrac{2011}{2012}\Leftrightarrow a=b=1\)

Thiếu đề k bn ???

\(A=\frac{x^2-64+5}{x+8}=x-8+\frac{5}{x-8}\)

Để A nguyên \(\Rightarrow5⋮\left(x-8\right)\Rightarrow x-8=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow x=\left\{3;5;9;13\right\}\)

\(B=\frac{x+2}{x^2+4}\)

Do \(x\in Z\Rightarrow\left|x+2\right|\le\left|x\right|+2\le x^2+2< x^2+4\)

\(\Rightarrow\frac{\left|x+2\right|}{x^2+4}< 1\Rightarrow-1< \frac{x+2}{x^2+4}< 1\)

\(\Rightarrow\frac{x+2}{x^2+4}=0\Rightarrow x=-2\)

ta có:

(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)

= (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)

= (\(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\)).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)

= \(\dfrac{6-x}{x-6}\)

= \(\dfrac{-\left(x-6\right)}{x-6}\)

= -1

Thêm điều kiện đề: a, b, c > 0

Giải:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)

Ta có a, b, c > 0 => a + b + c > 0

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow a=b=c\)

Vậy ...

Chúc bạn học tốt!

Đặt: \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)

\(\Leftrightarrow a^2+b^2+2bc+2ac+2c^2=0\)

\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)

\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\Leftrightarrow a=b=-c\Leftrightarrow x=y=-z\)

Giải tiếp nhé

\(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\left(2-\frac{1}{z}\right)^2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{z^2}=4-\frac{4}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}=-8\)

\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}-2=0\\\frac{1}{y}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\end{matrix}\right.\) \(\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=\frac{-1}{2}\)

Vậy \(P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2012}=1^{2012}=1\)

ĐK: \(x^3+x^2+x+1\ne0\Leftrightarrow x\ne-1\)

a) \(M=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\frac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\frac{3}{x^2+1}\)

b)Để M nhận giá trị nguyên thì \(x^2+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Mà \(x^2+1\ge1\forall x\) do đó \(x^2+1\in\left\{1;3\right\}\Leftrightarrow x^2\in\left\{0;2\right\}\Leftrightarrow x\in\left\{0;\sqrt{2}\right\}\)

Mà x nguyên nên x = 0

c) Do \(x^2+1\ge0+1=1\Rightarrow M=\frac{3}{x^2+1}\le\frac{3}{1}=3\)

Đẳng thức xảy ra khi x = 0

Vậy Max M = 3

P/s: Ko chắc câu b về cách trình bày..