Bài 36:

a.

Nếu $a,b>0$ thì:

\(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}.\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}}:\frac{a^2-b^2}{ab}\\ =\frac{a-b}{\sqrt{ab}}.\frac{ab}{(a-b)(a+b)}=\frac{\sqrt{ab}}{a+b}\)

Nếu $a,b<0$ thì:

\(A=\frac{\sqrt{-a}-\sqrt{-b}}{\sqrt{-b}}.\frac{\sqrt{-a}+\sqrt{-b}}{\sqrt{-a}}:\frac{a^2-b^2}{ab}\\ =\frac{(-a)-(-b)}{\sqrt{ab}}.\frac{ab}{(a-b)(a+b)}=\frac{\sqrt{ab}}{-(a+b)}\)

Vậy không có đáp án đúng.

b.

$b=1$ thì $b>0, a>0$.

Khi đó: $A=\frac{\sqrt{ab}}{a+b}=2$

$\Leftrightarrow \frac{\sqrt{a}}{a+1}=2$

$\Leftrightarrow \sqrt{a}=2a+2$

$\Leftrightarrow 2a-\sqrt{a}+2=0$

$\Leftrightarrow (\sqrt{a}-0,5)^2+a+1,75=0$

$\Leftrightarrow (\sqrt{a}-0,5)^2+a=-1,75<0$ (vô lý với mọi $a>0$)

Đáp án D.

\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2\sqrt{2}}{\sqrt{2}+1}-\left(3+\sqrt{3}-2\sqrt{2}\right)\\ =\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{2\sqrt{2}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-3-\sqrt{3}+2\sqrt{2}\\ =\sqrt{3}+2+\dfrac{4-2\sqrt{2}}{2-1}-3-\sqrt{3}+2\sqrt{2}\\ =-1+2\sqrt{2}+\dfrac{4-2\sqrt{2}}{1}\\ =-1+2\sqrt{2}+4-2\sqrt{2}\\ =3\)

Giúp vs mn ơi

Cái cuối là c(1/a+1/b) nha mn

1) \(\dfrac{\sqrt[]{6}+\sqrt[]{14}}{2\sqrt[]{3}+\sqrt[]{28}}=\dfrac{\sqrt[]{2}\left(\sqrt[]{3}+\sqrt[]{7}\right)}{2\left(\sqrt[]{3}+\sqrt[]{7}\right)}=\dfrac{\sqrt[]{2}}{2}\)

2) \(\dfrac{\sqrt[]{5}-\sqrt[]{10}}{\sqrt[]{3}-\sqrt[]{6}}=\dfrac{\sqrt[]{5}\left(1-\sqrt[]{2}\right)}{\sqrt[]{3}\left(1-\sqrt[]{2}\right)}=\sqrt[]{\dfrac{5}{3}}\)

3) \(...=\dfrac{4\sqrt[]{2}-2\sqrt[]{3}}{3\sqrt[]{2}-4\sqrt[]{3}}-\dfrac{\sqrt[]{5}+\sqrt[]{27}}{\sqrt[]{6}\left(\sqrt[]{5}+\sqrt[]{27}\right)}\)

\(=\dfrac{2\left(2\sqrt[]{2}-\sqrt[]{3}\right)}{3\sqrt[]{2}-4\sqrt[]{3}}-\dfrac{1}{\sqrt[]{6}}\)

\(=\dfrac{2\sqrt[]{6}\left(2\sqrt[]{2}-\sqrt[]{3}\right)-\left(3\sqrt[]{2}-4\sqrt[]{3}\right)}{\sqrt[]{6}\left(3\sqrt[]{2}-4\sqrt[]{3}\right)}\)

\(=\dfrac{8\sqrt[]{3}-6\sqrt[]{2}-3\sqrt[]{2}+4\sqrt[]{3}}{6\sqrt[]{3}-12\sqrt[]{3}}\)

\(=\dfrac{12\sqrt[]{3}-9\sqrt[]{2}}{-6\sqrt[]{3}}=-2+\sqrt[]{\dfrac{3}{2}}\)

Bài 4 bạn tự làm nhé

\(17-\dfrac{12}{\sqrt[]{2}}=17-6\sqrt[]{2}=18-6\sqrt[]{2}+1-2\)

\(=\left(\sqrt[]{18}-1\right)^2-2=\left(3\sqrt[]{2}-1\right)^2-2\)

Bài 1 :

\(...\Rightarrow A=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+2\right)+\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}-2\right)}{\left(\sqrt[]{x}-2\right)\left(\sqrt[]{x}+2\right)}-\dfrac{\sqrt[]{x}-10}{x-4}\)

\(\Rightarrow A=\dfrac{x+2\sqrt[]{x}+x-3\sqrt[]{x}+2}{x-4}-\dfrac{\sqrt[]{x}-10}{x-4}\)

\(\Rightarrow A=\dfrac{2x-\sqrt[]{x}+2-\sqrt[]{x}+10}{x-4}\)

\(\Rightarrow A=\dfrac{2x-2\sqrt[]{x}+12}{x-4}=\dfrac{2\left(x-\sqrt[]{x}+6\right)}{x-4}\)

Ta có:

\(P=\left(2+\sqrt{2}\right)^7+\left(2-\sqrt{2}\right)^7\)

\(P=2^7+7.2^6\sqrt{2}+21.2^5\left(\sqrt{2}\right)^2+...+7.2\left(\sqrt{2}\right)^6+\left(\sqrt{2}\right)^7\)\(+2^7-7.2^6\sqrt{2}+21.2^5\left(\sqrt{2}\right)^2-...+7.2\left(\sqrt{2}\right)^6-\left(\sqrt{2}\right)^7\)

\(P=2.2^7+2.21.2^5.\left(\sqrt{2}\right)^2+2.35.2^3.\left(\sqrt{2}\right)^4+2.7.2.\left(\sqrt{2}\right)^6\)

\(P=2^8+21.2^7+35.2^6+7.2^5\)

\(P=5408\)

\(\Rightarrow\left(2+\sqrt{2}\right)^7=5408-\left(2-\sqrt{2}\right)^7\)

Do \(0< \left(2-\sqrt{2}\right)^7< 1\) nên suy ra \(5047< \left(2+\sqrt{2}\right)^7< 5048\)

Vậy số nguyên lớn nhất không vượt quá \(\left(2+\sqrt{2}\right)^7\) là 5047.

(Sau này ta kí hiệu như thế này cho gọn.)

ĐKXĐ : \(x\inℝ\)

Ta có : x² + 4x + 7 = (x + 4)\(\sqrt{x^2+7}\)

\(\Leftrightarrow x^2+7+4x=x\sqrt{x^2+7}+4\sqrt{x^2+7}\) (*)

Đặt \(\sqrt{x^2+7}=a>0\)

Có (*) \(\Leftrightarrow a^2+4x=ax+4a\)

\(\Leftrightarrow\left(a-x\right).\left(a-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x\\a=4\end{matrix}\right.\)

Với a = x \(\Leftrightarrow\sqrt{x^2+7}=x\Leftrightarrow\left\{{}\begin{matrix}x^2+7=x^2\\x>0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Với a = 4 \(\Leftrightarrow\sqrt{x^2+7}=4\Leftrightarrow x^2+7=16\Leftrightarrow x=\pm3\)

Thử lại thấy thỏa mãn

Tập nghiệm \(S=\left\{\pm3\right\}\)

\(x^2+4x+7=\left(x+4\right)\sqrt[]{x^2+7}\)

\(\Leftrightarrow x^2+7+4x=\left(x+4\right)\sqrt[]{x^2+7}\left(1\right)\)

Đặt \(t=\sqrt[]{x^2+7}\left(t\ge0\right)\)

\(\left(1\right)\Leftrightarrow t^2+4x=\left(x+4\right).t\)

\(\Leftrightarrow t^2-\left(x+4\right).t+4x=0\)

\(\Leftrightarrow t^2-tx-4.t+4x=0\)

\(\Leftrightarrow t\left(t-x\right)-4\left(t-x\right)=0\)

\(\Leftrightarrow\left(t-x\right)\left(t-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-x=0\\t-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t=x\\t=4\end{matrix}\right.\)

- Với \(t=x\Leftrightarrow\sqrt[]{x^2+7}=x\Leftrightarrow x^2+7=x^2\Leftrightarrow0.x^7=7\left(loại\right)\)

- Với \(t=4\Leftrightarrow\sqrt[]{x^2+7}=4\Leftrightarrow x^2+7=16\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)

Vậy nghiệm phương trình là \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Dễ dàng chứng minh \(OM.OC=OA^2=OD^2\). \(\Rightarrow\Delta OMD~\Delta ODC\left(c.g.c\right)\) \(\Rightarrow\widehat{ODM}=\widehat{OCD}\)

Mà \(\widehat{OCD}=\widehat{ADC}\) (do OC//AD (\(\perp AB\))) nên suy ra \(\widehat{ADC}=\widehat{BDM}\) (đpcm)