Ôn thi vào 10

Vì `x>0` nên ta chia 2 vế tử và mẫu cho `sqrtx>0`

`=>sqrx/(x-sqrtx+1)`

`=1/(sqrtx-1+1/sqrtx)`

Áp dụng cosi:

`sqrtx+1/sqrtx>=2`

`=>sqrtx-1+1/sqrtx>=1`

`=>1/(sqrtx-1+1/sqrtx)<=1`

Hay `sqrtx/(x-sqrtx+1)<=1`

Dấu "=" `<=>x=1`

`A=(sqrtx+3)/(sqrtx-4)-(sqrtx+2)/(sqrtx-3)-(5(sqrtx-2))/(x-7sqrtx+12)(x ne 9,16)`

`=(x-9-(x-2sqrtx-8)-5sqrtx+10)/(x-7sqrtx+12)`

`=(x-9-x+2sqrtx+8-5sqrtx+10)/(x-7sqrtx+12)`

`=(-3sqrtx+9)/(x-7sqrtx+12)`

`=(-3(sqrtx-3))/((sqrtx-3)(sqrtx-4))`

`=-3/(sqrtx-4)`

Câu 1:

a) Khi x =16 (t.m ĐKXĐ) thì B có giá trị là:

\(B=\dfrac{16-6\cdot4}{4-1}=\dfrac{-8}{3}\)

b) Ta có:

\(A=\dfrac{25\sqrt{x}+6}{x-36}-\dfrac{\sqrt{x}-1}{6-\sqrt{x}}+\dfrac{2\sqrt{x}}{\sqrt{x}+6}=\dfrac{25\sqrt{x}+6}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{25\sqrt{x}+6+x+5\sqrt{x}-6+2x-12\sqrt{x}}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{3x+18\sqrt{x}}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-6}\)

c) Ta có:

\(T=\sqrt{A\cdot B}=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{x-6\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3x\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}-1\right)}}=\sqrt{\dfrac{3\left(x-1\right)+3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}+1\right)+\dfrac{3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\right)+6}\overset{Cosi}{\ge}\sqrt{3\cdot2+6}=2\sqrt{3}\)

Dấu = xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(t.m\right)\)

=20,39819798

`sqrt{9-4sqrt5}+3sqrt5+11`

`=sqrt{5-2.2.sqrt5+4}+3sqrt5+11`

`=sqrt{(sqrt5-2)^2}+3sqrt5+11`

`=sqrt5-2+3sqrt5+11`

`=9+4sqrt5`

\(\sqrt{9-4\sqrt 5}+3\sqrt 5+11\\=\sqrt{4-4\sqrt 5+5}+3\sqrt 5+11\\=\sqrt{(2-\sqrt 5)^2}+3\sqrt 5+11\\=|2-\sqrt 5|+3\sqrt 5+11\\=\sqrt 5-2+3\sqrt 5+11\\=4\sqrt 5+9\)

\(x\ge xy+1\Rightarrow1\ge y+\dfrac{1}{x}\ge2\sqrt{\dfrac{y}{x}}\Rightarrow\dfrac{y}{x}\le\dfrac{1}{4}\)

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

Đặt \(\dfrac{y}{x}=t\le\dfrac{1}{4}\) 

\(Q^2=\dfrac{t^2+2t+1}{t^2-t+3}=\dfrac{t^2+2t+1}{t^2-t+3}-\dfrac{5}{9}+\dfrac{5}{9}\)

\(Q^2=\dfrac{\left(4t-1\right)\left(t+6\right)}{9\left(t^2-t+3\right)}+\dfrac{5}{9}\le\dfrac{5}{9}\)

\(\Rightarrow Q_{max}=\dfrac{\sqrt{5}}{3}\) khi \(t=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(2;\dfrac{1}{2}\right)\)

ĐK: x > 0

B = \(\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

⇔ B = \(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

⇔ B = \(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\) = \(\dfrac{2}{x-1}\)

Để B ∈ Z thì x - 1 ∈ Ư(2) = {-2;-1;1;2}

⇔ \(\left\{{}\begin{matrix}x-1=-2\\x-1=-1\\x-1=1\\x-1=2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=-3\\x=-2\\x=2\\x=3\end{matrix}\right.\)

Vậy....