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\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)
\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)
\(P>0\Leftrightarrow x>-4\)
7,
\(\Leftrightarrow x=\sqrt{x+2}\left(\frac{\sqrt{x}}{1+\sqrt{1-\sqrt{x}}}\right)^2\)
\(\Leftrightarrow x=\frac{\sqrt{x+2}.x}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}\Leftrightarrow\frac{\sqrt{x+2}}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}=1\)
đến đây tự làm
7 đề như tớ
8. (x-1)^2 +\(x\sqrt{x-\frac{1}{x}}\)
9. \(\sqrt{1+x}+\sqrt{3-3x}=\sqrt{4x^2+1}\)
30. \(\tan x+\cot x=2\sin\left(x+\frac{\pi}{4}\right)\)
ĐK: \(x\ne\frac{k\pi}{2}\)
pt <=> \(\frac{1}{\sin x.\cos x}=2\sin\left(x+\frac{\pi}{4}\right)\)
<=> \(\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)\)
Đánh giá: \(-1\le\sin2x\le1\)
=> \(\orbr{\begin{cases}\frac{1}{\sin2x}\le-1\\\frac{1}{\sin2x}\ge1\end{cases}}\)
\(-1\le\sin\left(x+\frac{\pi}{4}\right)\le1\)
Như vậy dấu "=" xảy ra <=> \(\orbr{\begin{cases}\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)=-1\\\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}\)
<=> \(\orbr{\begin{cases}\sin2x=\sin\left(x+\frac{\pi}{4}\right)=-1\\\sin2x=\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}\)
TH1: \(\sin2x=\sin\left(x+\frac{\pi}{4}\right)=-1\)
<=> \(\hept{\begin{cases}2x=-\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{\pi}{4}+k\pi\\x=-\frac{3\pi}{4}+k2\pi\end{cases}}\)loại
TH2:
\(\sin2x=\sin\left(x+\frac{\pi}{4}\right)=1\)
<=> \(\hept{\begin{cases}2x=\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{4}+k2\pi\end{cases}}\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)
Vậy ...
29) \(\sin x-2\sin2x-\sin3x=2\sqrt{2}\)
<=> \(\left(\sin x-\sin3x\right)-2\sin2x=2\sqrt{2}\)
<=> \(-2.\sin x\cos2x-2\sin2x=2\sqrt{2}\)
<=> \(\sin x\cos2x+\sin2x=-\sqrt{2}\)
Ta có: \(\left(\sin x\cos2x+\sin2x\right)^2\le\left(\sin^2x+1\right)\left(\sin^22x+\cos^22x\right)=\sin^2x+1\le2\)
( theo bunhia)
=> \(-\sqrt{2}\le\sin x\cos2x+\sin2x\le\sqrt{2}\)
Dấu "=" xảy ra <=> \(\frac{\sin x}{1}=\frac{\cos2x}{\sin2x}\)(1) và \(\sin x\cos2x+\sin2x=-\sqrt{2}\)(2)
(1) <=> \(\frac{\sin x.\cos2x}{1}=\frac{\cos^22x}{\sin2x}\)=> (2) <=> \(\frac{\cos^22x}{\sin2x}+\sin2x=-\sqrt{2}\)
<=> \(\frac{1}{\sin2x}=-\sqrt{2}\)<=> \(\sin2x=-\frac{\sqrt{2}}{2}\)<=> \(\orbr{\begin{cases}x=-\frac{\pi}{8}+k\pi\\x=-\frac{3\pi}{8}+k\pi\end{cases}}\)
(1) <=> \(\sin x.\sin2x=\cos2x\)=> (2) <=> \(\sin x.\sin x.\sin2x+\sin2x=-\sqrt{2}\)
<=> \(\frac{\sin^2x}{2}+\frac{1}{2}=+1\Leftrightarrow\sin^2x=1\)=> \(\cos^2x=0\)loại vì \(\sin2x=-\frac{\sqrt{2}}{2}\)
Vậy pt vô nghiệm