Căn bậc hai

Ta có: \(\sqrt{\left(11-6\sqrt{2}\right)^2}\)

\(=\left|11-6\sqrt{2}\right|\)

\(=11-6\sqrt{2}\)

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1-4}{\sqrt{x+1}}\\ \dfrac{\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{4}{\sqrt{x}+1}\\ =1-\dfrac{4}{\sqrt{x}+1}\\A\in Z\\ \Rightarrow1-\dfrac{4}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\left(\sqrt{x}+1\right)\inƯ\left(4\right)\\ \Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-3;-2;0;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;1;3\right\}\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x\in\left\{0;1;9\right\}\)

Vậy \(A\) nguyên \(\Leftrightarrow x\in\left\{0;1;9\right\}\)

Ta có bài toán quen thuộc sau: 

Nếu \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\) thì \(x+y=0\)

Do đó từ giả thiết ta chỉ cần chứng minh được \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\) thì bài toán được giải quyết.

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+1=a^2+x^2-2ax\\y^2+1=b^2+y^2-2by\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

Thế vào giả thiết:

\(\left(\dfrac{a^2-1}{2a}+\sqrt{1+\left(\dfrac{b^2-1}{2b}\right)^2}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{1+\left(\dfrac{a^2-1}{2a}\right)^2}\right)=1\)

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

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

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

Chú ý rằng: \(1=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4ab}\)

Do đó (1) tương đương:

\(\left(\dfrac{a+b}{2}\right)^2-\dfrac{\left(a-b\right)^2}{\left(2ab\right)^2}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow\left[\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right]\left(1-\dfrac{1}{ab}\right)=0\)

Do \(a;b>0\Rightarrow\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}>0\)

\(\Rightarrow1-\dfrac{1}{ab}=0\Leftrightarrow ab=1\)

Hay \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Rightarrow x+y=0\Rightarrow P=100\)

Tìm điều gì của M bạn?

Mình nghĩ là tìm Min, Max \(M=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\).

Tìm Min: Ta có \(M^2\ge a+b+b+c+c+a=2\left(a+b+c\right)\ge2\sqrt{a^2+b^2+c^2}=2\).

Do đó \(M\geq\sqrt{2}\).Đẳng thức xảy ra khi a = b = 0; c = 1.

Tìm Max: Ta có \(M\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}=\sqrt[4]{108}\).

Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{2015-2014}{\sqrt{2015}+\sqrt{2014}}>\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\sqrt{2016}-\sqrt{2015}\)

Ta có:  √2015−√2014=2015−2014√2015+√2014>2016−2015√2016+√2015=√2016−√2015

a, b bạn tự giải

c. \(\Delta=m^2+4>0;\forall m\Rightarrow\) pt luôn có nghiệm

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-1\end{matrix}\right.\)

Ồ, đề câu d bạn ghi sai, 2 mẫu số phải có 1 cái là \(x_1\)

\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\dfrac{3\sqrt{5}-5}{1}+\dfrac{5+\sqrt{5}}{4}-\dfrac{9\sqrt{5}-15}{4}\)

\(=\sqrt{5}\)