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     Roots in the complex plane:

a/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\ge0\\\sqrt{1+x}=b\ge0\end{matrix}\right.\) được hệ:

\(\left\{{}\begin{matrix}\sqrt{1+ab}\left(a^3-b^3\right)=2+ab\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)\left(a^2+ab+b^2\right)=a^2+b^2+ab\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)=1\\a^2+b^2=2\end{matrix}\right.\) \(\left(a\ge b\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(a-b\right)^2=1\\a^2+b^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(2-2ab\right)=1\\a^2+b^2=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}1-a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\)

Theo Viet đảo, \(a^2;b^2\) là nghiệm của:

\(t^2-2t+\frac{1}{2}=0\Rightarrow\left[{}\begin{matrix}t=\frac{2+\sqrt{2}}{2}\\t=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\)

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

2 phần còn lại ko biết giải theo kiểu lớp 10, chỉ biết lượng giác hóa, bạn tham khảo thôi :(

b/ Đặt \(x=cos2t\) pt trở thành:

\(\sqrt{1-cos2t}-2cos2t.\sqrt{1-cos^22t}-\left(2cos^22t-1\right)=0\)

\(\Leftrightarrow\sqrt{2}sint-2sin2t.cos2t-cos4t=0\)

\(\Leftrightarrow\sqrt{2}sint-sin4t-cos4t=0\)

\(\Leftrightarrow\sqrt{2}sint=sin4t+cos4t=\sqrt{2}sin\left(4t+\frac{\pi}{4}\right)\)

\(\Leftrightarrow sin\left(4t+\frac{\pi}{4}\right)=sint\)

\(\Leftrightarrow\left[{}\begin{matrix}4t+\frac{\pi}{4}=t+k2\pi\\4t+\frac{\pi}{4}=\pi-t+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\frac{\pi}{12}+\frac{k2\pi}{3}\\t=-\frac{\pi}{20}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=cos\left(-\frac{\pi}{6}+\frac{k4\pi}{3}\right)\\x=cos\left(-\frac{\pi}{10}+\frac{k4\pi}{5}\right)\end{matrix}\right.\) với \(k\in Z\)

c/ Đặt \(x=cost\)

\(64cos^6t-112cos^4t+56cos^2t-7=2\sqrt{1-cos^2t}\)

\(\Leftrightarrow64cos^6t-112cos^4t+56cos^2t-7=2sint\)

Nhận thấy \(cost=0\) không phải nghiệm, pt tương đương:

\(64cos^7t-112cos^5t+56cos^3t-7cost=2sint.cost\)

\(\Leftrightarrow cos7t=sin2t=cos\left(\frac{\pi}{2}-2t\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7t=\frac{\pi}{2}-2t+k2\pi\\7t=2t-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{18}+\frac{k2\pi}{9}\\t=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=cos\left(\frac{\pi}{18}+\frac{k2\pi}{9}\right)\\x=\left(-\frac{\pi}{10}+\frac{k2\pi}{5}\right)\end{matrix}\right.\)

Ý tưởng của người ra đề khá kì quặc, công thức \(cos7a\) kia thực sự là chứng minh rất mất thời gian

Đúng đề chưa vậy

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

 

a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)

Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)

    \(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)

     \(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)

     \(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)

     \(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)

     \(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)

Vậy \(S=\left\{1\right\}\)

     

x= 0.761322463768116,

x= 0.369494467346496,

x=1.57660410301179