Bài 9: Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

x,y,z ở đâu ra vậy

ta có x-6y=xy

=) x-2y=5xy

=)x+2y=9xy

suy ra M=\(\dfrac{x-2y}{x+2y}=\dfrac{5xy}{9xy}=\dfrac{5}{9}\)

ko biết đến đây đã được chưa :V

\(M=\dfrac{x-2y}{x+2y}\\ =\dfrac{x-6y+4y}{x-6y+8y}\\ =\dfrac{xy+4xy}{xy+8xy}\\ =\dfrac{y\left(x+4\right)}{y\left(x+8\right)}\\ =\dfrac{x+4}{x+8}\)

thế đó, chúc may mắn :)

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{z+y+z}=9=\dfrac{18}{2}>\dfrac{18}{xyz+2}\)

Ta có: x²-6y²=xy

<=> x²-xy-6y² =0

<=> x²-3xy+2xy-6y²=0

<=> x(x-3y)+2y(x-3y)=0

<=>(x+2y)(x-3y)=0

+ Với x+2y = 0 <=> x=-2y, thay vào M ta được:

M=-2y-2y(-2y)+2y

= 4y²

+ Với x-3y =0 <=> x=3y, thay vào M ta được:

M= 3y-2y.3y+2y

= 5y-6y²

= y(5-6y)

Dùng phương pháp biến đổi tương đương nhé!!!

Ta có : \(\dfrac{1}{1+a^2}\) + \(\dfrac{1}{1+b^2}\) \(\ge\) \(\dfrac{2}{1+ab}\)

<=>( \(\dfrac{1}{1+a^2}\) - \(\dfrac{1}{1+ab}\) ) + ( \(\dfrac{1}{1+b^2}\) - \(\dfrac{1}{1+ab}\) ) \(\ge\) 0

<=> \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\) \(\ge\) 0

<=> \(\left(b-a\right)\left(a+ab^2-b-a^2b\right)\) \(\ge\) 0

<=> \(\left(b-a\right)\left[ab\left(b-a\right)-\left(b-a\right)\right]\) \(\ge\) 0

<=> \(\left(b-a\right)\left(b-a\right)\left(ab-1\right)\) \(\ge\) 0

<=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0 (1)

Mà \(\left\{{}\begin{matrix}\left(b-a\right)^2\ge0\\ab-1\ge0\end{matrix}\right.\) ( vì ab \(\ge\)1)

=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0

=> (1) luôn đúng

Vậy đpcm ....

Ta có: \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

\(\Leftrightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+b^2}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

BĐT cuối cùng đúng vì \(a.b\ge1\Rightarrowđpcm\)

ta có: \(A=\dfrac{x^2}{\left(x^2+1\right)^3}=\dfrac{x^2}{x^6+3x^4+3x^2+1}=\dfrac{1}{x^4+3x^2+3+\dfrac{1}{x^2}}\)

đặt \(x^2=a\left(a\ge0\right)\Rightarrow A=\dfrac{1}{a^2+3a+3+\dfrac{1}{a}}\)

ta đi tìm min của \(P=a^2+3a+3+\dfrac{1}{a}=a^2-a+4a+\dfrac{1}{a}+3\)

\(=\left(a^2-a+\dfrac{1}{4}\right)+\left(4a+\dfrac{1}{a}\right)+\dfrac{11}{4}=\left(a-\dfrac{1}{2}\right)^2+\left(4a+\dfrac{1}{a}\right)+\dfrac{11}{4}\)

a >0;Áp dụng BĐT cauchy: \(4a+\dfrac{1}{a}\ge2\sqrt{4a.\dfrac{1}{a}}=4\)

do đó \(P\ge4+\dfrac{11}{4}=\dfrac{27}{4}\)( vì \(\left(a-\dfrac{1}{2}\right)^2\ge0\))

\(\Rightarrow A\le\dfrac{4}{27}\)

dấu = xảy ra khi \(4a=\dfrac{1}{a}\Leftrightarrow a=\dfrac{1}{2}\left(a\ge0\right)\)và nó cũng trùng với \(\left(a-\dfrac{1}{2}\right)^2\ge0\)

khi đó \(x=\pm\dfrac{\sqrt{2}}{2}\)

\(D=\dfrac{x^2-3x+3}{x^2-2x+1}=\dfrac{3}{4}+\dfrac{x^2-3x+3-\dfrac{3}{4}x^2+\dfrac{9}{4}x-\dfrac{9}{4}}{x^2-2x+1}=\dfrac{3}{4}+\dfrac{\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{3}{4}}{x^2-2x+1}\)\(=\dfrac{3}{4}+\dfrac{\left(\dfrac{1}{2}x-\sqrt{\dfrac{3}{4}}\right)^2}{\left(x-1\right)^2}\le\dfrac{3}{4}\)

Vậy \(Min_D=\dfrac{3}{4}\)khi \(\dfrac{1}{2}x-\sqrt{\dfrac{3}{4}}=0\Rightarrow\dfrac{1}{2}x=\sqrt{\dfrac{3}{4}}\Rightarrow x=\sqrt{\dfrac{3}{4}}-\dfrac{1}{2}\)

\(A=x^2-4xy+2x-4y+3+4y^2\)

\(A=x^2-2.2xy+\left(2y\right)^2+2x-4y+3\)

\(A=\left(x-2y\right)^2-2.\left(x-2y\right)+1+2\)

\(A=\left(x-2y-1\right)^2+2\ge2\)

Vậy GTNN của A=2.