Bài 3: Rút gọn phân thức

\(Q=\left(\dfrac{1}{x+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2}{x^2-x+1}\right):\left(x+2\right)\)\(=\left(\dfrac{x^2-x+1}{x^3+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2\left(x+1\right)}{x^3+1}\right).\dfrac{1}{x+2}\)\(=\left(\dfrac{x^2-x+1+6x+3-2x-2}{\left(x+1\right)\left(x-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\left(\dfrac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{1}{x+2}=\dfrac{1}{\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)Ta có:

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) \(\Rightarrow\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

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

= \(\dfrac{2+x}{2-x}\)- \(\dfrac{4x^2}{^{^{ }}4-x^2}\)- \(\dfrac{2-x}{2+x}\): \(\dfrac{x^2-3x}{2x^2-x^3}\)

= \(\dfrac{\left(2+x\right)\left(2+x\right)-4x^{2^{ }}+\left(2-x\right)\left(2-x\right)}{4-x^2}\) . \(\dfrac{2x^{2^{ }}-x^3}{x^{2^{ }}-3x}\)

= \(\dfrac{4+4x+x^{2^{ }}-4x^{2^{ }}-4+4x-x^2}{^{ }4-x^2}\). \(\dfrac{2x^{2^{ }}-x^3}{x^2-3x}\)

= \(\dfrac{8x-4x^2}{4-x^2}\). \(\dfrac{2x^2-x^3}{x^2-3x}\)

= \(\dfrac{8-4x^2}{2+x}\). \(\dfrac{x^2}{x-3}\)

mình viết nhầm hàng thứ 2 phải là -(2-x)(2-x)

= - - :

= .

= .

= .

= .

Ta có:

\(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}=\dfrac{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}-\dfrac{x-9}{6-x-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}\)\(=\dfrac{x-9}{6-x-\sqrt{x}}-\dfrac{x-9}{6-x-\sqrt{x}}+\dfrac{\sqrt{x}+2}{3+\sqrt{x}}=\dfrac{\sqrt{x}+2}{3+\sqrt{x}}\)(1)

\(1-\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{x-9-x-3\sqrt{x}}{\left(\sqrt{x}\right)^2-3^2}=\dfrac{-3\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)}=\dfrac{-3}{\sqrt{x}-3}\left(2\right)\)Thay (1) và (2) vào biểu thức ta được

\(\dfrac{-3}{\sqrt{x}-3}:\dfrac{\sqrt{x}-2}{3+\sqrt{x}}=\dfrac{-3\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

Cái này mình không chắc lắm , không biết còn rút gọn được không nữa!

Ta có: A = \(\dfrac{27-12x}{x^2+9}\) = \(\dfrac{\left(4x^2+36\right)-\left(4x^2+12x+9\right)}{x^2+9}\)

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

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

Vì \(\left(2x+3\right)^2\) \(\ge\) 0

\(x^2+9\) > 0

=> \(\dfrac{\left(2x+3\right)^2}{x^2+9}\) \(\ge\) 0

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

Dấu bằng xảy ra <=> \(\left(2x+3\right)^2\) = 0

<=> 2x +3 = 0

<=> x = \(\dfrac{-3}{2}\)

Vậy GTLN của A = 4 khi x = \(\dfrac{-3}{2}\)

câu này trên google

bạn nên tra google trước khi đăng

\(\dfrac{4X\left(X^2-4X+4\right)}{\left(X-2\right)\left(X+2\right)}\)= \(\dfrac{4X\left(X-2\right)^2}{\left(X-2\right)\left(X+2\right)}\)= \(\dfrac{4X\left(X-2\right)}{X+2}\)

Khi x<2 , ta có \(x-2< 0\) nên | x - 2 | = -(x - 2) = -x + 2

Vậy M = 3x - x + 2 = 2x + 2

Xét tam giác AKC và AHB có:

\(\widehat{A}\) là góc chung

\(\widehat{AKC}=\widehat{AHB}\) (gt)

\(\Rightarrow\) Tam giác AKC đồng dạng với tam giác AHB(g.g)

\(\Rightarrow\dfrac{AH}{AC}=\dfrac{AK}{AB}\Rightarrow AH.AB=AC.AK\)

Xét ΔvgHAB và ΔvgKAC

có \(\widehat{A}\) chung

⇒ΔvgHAB ~ ΔvgKAC

⇒\(\dfrac{AH}{AK}=\dfrac{AB}{AC}\)

\(\Rightarrow AH.AC=AK.AB\)