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)