Đặt \(\sqrt{x^2+3}=a\ge\sqrt{3}\) (1)

pt \(\Leftrightarrow\left(a^2-3\right)^2+a-3=0\)

\(\Leftrightarrow a^4+9-6a^2+a-3=0\)

\(\Leftrightarrow a^4-4a^2-2a^2+4a-3a+6=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2-2a-3=0\right)\)

\(\Leftrightarrow\left(a-2\right)\left(a^3+a^2+a^2+a-3a-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left(a^2+a-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left[\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}a-2=0\\a+1=0\\\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(c\right)\\a=-1\left(l\right)\\a=\dfrac{-1+\sqrt{13}}{2}\left(l\right)\\a=\dfrac{-1-\sqrt{13}}{2}\left(l\right)\end{matrix}\right.\)

Thay a = 2 vào (1) ta được: \(\sqrt{x^2+3}=2\Rightarrow x^2+3=4\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

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

\(=\sqrt{x^2+1}+\dfrac{1}{\sqrt{x^2+1}}\ge2\sqrt{\sqrt{x^2+1}.\dfrac{1}{\sqrt{x^2+1}}}=2\)

Dấu "=" xảy ra khi x = 0

\(2016^{2017}=\left(2^5.3^2.7\right)^{2017}=2^{10085}.3^{4034}.7^{2017}\)

Ta có: \(2^{20}\equiv100\left(mod100\right)\)\(\Rightarrow2^{10085}=\left(2^{20}\right)^{504}.2^5\equiv76.32\left(mod100\right)\equiv32\left(mod100\right)\)

\(3^{20}\equiv1\left(mod100\right)\)\(\Rightarrow3^{4034}=\left(3^{20}\right)^{201}.3^{14}\equiv1.\left(...69\right)\equiv69\left(mod100\right)\)

\(7^4\equiv1\left(mod100\right)\)\(\Rightarrow7^{2017}=\left(7^4\right)^{504}.7\equiv1.7\left(mod100\right)\equiv7\left(mod100\right)\)

Như vậy \(2016^{2017}\equiv32.69.7\left(mod100\right)\equiv56\left(mod100\right)\)

Vậy 2 chữ số tận cùng của 2016²⁰¹⁷ là 56

(x - 2)² - 3(x + 1)² = 2 - (x - 2)(x - 1)

<=> (x - 2)² + (x - 2)(x - 1) = 2 + 3(x + 1)²

<=> (x - 2)(x - 2 + x - 1) = 2 + 3(x² + 1 + 2x)

<=> (x - 2).(2x - 3) = 2 + 3x² + 3 + 6x

<=> 2x² - 4x - 3x + 6 = 2 + 3x² + 3 + 6x

<=> x² + 13x - 1 = 0

<=> \(\left(x+\dfrac{13}{2}\right)^2=\dfrac{173}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13+\sqrt{173}}{2}\\x=\dfrac{-13-\sqrt{173}}{2}\end{matrix}\right.\)

Vậy ...

a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{6}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{6}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

b) Tương tự

Số hạng thứ k+1 trong khai triển nhị thức 2 + x n  là 

A.  C n k 2 n x k

B. C n k 2 n - k x k

C. C n k 2 n - k x n

D. C n k + 1 2 n - k - 1 x k + 1

a) \(\Delta DMC=\Delta AMB\left(c.g.c\right)\)

=> góc DCM = góc ABM (2 góc tương ứng)

Mà 2 góc này ở vị trí so le trong nên DC // AB (đpcm)

b) cần chứng minh thêm E, M, F thẳng hàng

c) gần TT câu a

ĐK: ...

\(t=a\sqrt{\dfrac{x^2+1}{2x}}\)\(\Rightarrow t^2=a^2.\dfrac{x^2+1}{2x}\)

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

Thay vào M ta được:

\(M=\left(\dfrac{\sqrt{a^2.\dfrac{\left(x-1\right)^2}{2x}}+\sqrt{a^2.\dfrac{\left(x+1\right)^2}{2x}}}{\sqrt{a^2.\dfrac{\left(x-1\right)^2}{2x}}-\sqrt{a^2.\dfrac{\left(x+1\right)^2}{2x}}}\right)^4\)

\(M=\left(\dfrac{\dfrac{a.\left(x-1\right)}{\sqrt{2x}}+\dfrac{a.\left(x+1\right)}{\sqrt{2x}}}{\dfrac{a.\left(x-1\right)}{\sqrt{2x}}-\dfrac{a.\left(x+1\right)}{\sqrt{2x}}}\right)^4\)

\(M=\left(\dfrac{x-1+x+1}{x-1-\left(x+1\right)}\right)^4=\left(\dfrac{2x}{-2}\right)^4=\left(-x\right)^4=x^4=2012^4\)