IR IRAN(Islamic Republic of Iran)
Giới thiệu về bản thân
a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)
Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)
Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)
\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)
\(\Leftrightarrow b=a\)
Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)
\(\Leftrightarrow x^3-4x^2-6x+5=0\)
\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)
ĐKXĐ : \(x\ge-4\)
(2x + 1)2 = (x + 4)\(\sqrt{4x^2+1}\)
<=> \(4x^2+1+4x=x\sqrt{4x^2+1}+4\sqrt{4x^2+1}\)
<=> \(\left(\sqrt{4x^2+1}-4\right)\left(\sqrt{4x^2+1}-x\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{4x^2+1}=4\\\sqrt{4x^2+1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2=15\\\left\{{}\begin{matrix}3x^2+1=0\\x\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\sqrt{15}}{2}\left(tm\right)\\∄x\end{matrix}\right.\Leftrightarrow x=\pm\dfrac{\sqrt{15}}{2}\)
B = (sin2a + cos2a)2 = 12 = 1
Đặt x25 = t
=> x100 + x50 + 1 = t4 + t2 + 1
= t4 + 2t2 + 1 - t2
= (t2 + 1)2 - t2
= (t2 - t + 1)(t2 + t + 1)
= (x50 - x25 + 1)(x50 + x25 + 1)