\(5x^2-x+5=\sqrt{x^4+x^2+1}\)

\(\Leftrightarrow5x^2-x+5=\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Đặt \(a=\sqrt{x^2-x+1};b=\sqrt{x^2+x+1}\left(a;b>0\right)\)

Pt tt: \(3a^2+2b^2=ab\)

\(\Leftrightarrow3a^2-ab+2b^2=0\) 

\(\Leftrightarrow3\left(a-\dfrac{b}{6}\right)^2+\dfrac{23}{12}b^2=0\)(vô nghiệm)

Vậy pt vô nghiệm

15C

16C

17B

18A

19A

20C

Câu 89:

\(cos^24x+cos^28x=sin^212x+sin^216x+2\)

\(\Leftrightarrow cos^24x+cos^28x=sin^212x+sin^216x+sin^24x+cos^24x+sin^28x+cos^28x\)

\(\Leftrightarrow sin^212x+sin^216x+sin^24x+sin^28x=0\)

Có \(VT\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}sin12x=0\\sin16x=0\\sin4x=0\\sin8x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\pi}{12}\\x=\dfrac{k\pi}{16}\\x=\dfrac{k\pi}{4}\\x=\dfrac{k\pi}{8}\end{matrix}\right.\)\(\Rightarrow x=0\)

Vậy x=0

Câu 88:

\(\sqrt{5+sin^2x}=sinx+2cosx\)

\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=\dfrac{1}{\sqrt{5}}sinx+\dfrac{2}{\sqrt{5}}cosx\)

\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\)

Có \(\sqrt{1+\dfrac{sin^2x}{5}}\ge\sqrt{1+0}=1\)

\(sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\le1\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=0\\sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x+arc.cos\dfrac{1}{\sqrt{5}}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x=k\pi-arc.cos\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy \(S=\varnothing\)

\(2S=3^{31}-1=3^{28}.3^3-1=\left(...1\right).27-1=\left(.....7\right)-1=\left(...6\right)\)

\(\Rightarrow S=\left(...3\right)\)

Tận cùng bằng 3 nhé e

Để căn thức có nghĩa\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1\le0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow x+1< 0\Leftrightarrow x< -1\)

Vậy...