\(\Leftrightarrow16x^4-4x^2-4xy+y^2+1=0\)

\(\Leftrightarrow\left(16x^4-8x^2+1\right)+\left(4x^2-4xy+y^2\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)^2+\left(2x-y\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-1=0\\2x-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(-\dfrac{1}{2};-1\right);\left(\dfrac{1}{2};1\right)\)

\(5.a.V_{rượu}=\dfrac{46.25}{100}=11,5\left(l\right)\\ m_{rượu}=11,5.0,8=9,2\left(g\right)\\ b.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,2\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,2.88=17,6\left(g\right)\\ VìH=30\%\Rightarrow m_{CH_3COOC_2H_5}=17,6.30\%=5,28\left(g\right)\)

\(6.a.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ n_{CH_3COOH}=0,5.60=30\left(g\right)\\ b.n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,5\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,5.88=44\left(g\right)\\ VìH=70\%\Rightarrow m_{CH_3COOC_2H_5}=44.70\%=30,8\left(g\right)\)

H₂ + CuO \(\xrightarrow[]{t^o}\) Cu + H₂O

a) nCuO = 16 : 80 = 0,2mol

Theo pt: nH₂ = nCuO = 0,2 mol

=> V H₂ = 0,2.22,4 = 4,48 lít

b) Theo pt: nCu = nCuO = 0,2 mol

=> mCu = 0,2 . 64 = 12,8g

nH₂O = nCuO = 0,2 mol

=> mH₂O = 0,2.18 = 3,6g

c) Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

Theo pt: nFe = nH₂ = 0,2 mol

=> mFe = 0,2.56 = 11,2g

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2315

13: \(cos^2x-\sqrt3\cdot\sin2x=1+\sin^2x\)

=>\(cos^2x-\sin^2x-\sqrt3\cdot\sin2x=1\)

=>\(cos2x-\sqrt3\cdot\sin2x=1\)

=>\(\frac12\cdot cos2x-\frac{\sqrt3}{2}\cdot\sin2x=\frac12\)

=>\(\sin\left(\frac{\pi}{6}-2x\right)=\frac12\)

=>\(\sin\left(2x-\frac{\pi}{6}\right)=-\frac12\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{6}=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=k2\pi\\ 2x=\frac43\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=k\pi\\ x=\frac23\pi+k\pi\end{array}\right.\)

12: \(\sin2x-2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot cosx-2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\left(cosx-\sin x\right)=0\)

=>\(\sin x\cdot\left(\cos x-\sin x\right)=0\)

TH1: sin x=0

=>\(x=k\pi\)

TH2: cos x-sin x=0

=>\(\sin x=cosx\)

=>\(\sin x=\sin\left(\frac{\pi}{2}-x\right)\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{2}-x+k2\pi\\ x=\pi-\frac{\pi}{2}+x+k2\pi=\frac{\pi}{2}+x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{2}+k2\pi\\ \frac{\pi}{2}+k2\pi=0\left(vôlý\right)\end{array}\right.\)

=>\(x=\frac{\pi}{4}+k\pi\)

11: \(\sin2x+2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot cosx+2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot\left(cosx+\sin x\right)=0\)

=>\(\sin x\cdot\left(cosx+\sin x\right)=0\)

TH1: \(\sin x=0\)

=>\(x=k\pi\)

TH2: sin x+cos x=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(x+\frac{\pi}{4}=k\pi\)

=>\(x=k\pi-\frac{\pi}{4}\)

5: \(2\cdot cos^2x-2\cdot cos2x-4\cdot\sin x\cdot cosx+1=0\)

=>\(2\cdot cos^2x-1-2\cdot cos2x-2\cdot\sin2x+2=0\)

=>\(cos2x-2\cdot cos2x-2\cdot\sin2x+2=0\)

=>\(-cos2x-2\cdot\sin2x=-2\)

=>\(2\cdot\sin2x+cos2x=2\)

=>\(\sin2x\cdot\frac{2}{\sqrt5}+cos2x\cdot\frac{1}{\sqrt5}=\frac{2}{\sqrt5}\)

=>\(\sin\left(2x+\alpha\right)=cos\left(\alpha\right)=\sin\left(\frac{\pi}{2}-\alpha\right)\)

=>\(\left[\begin{array}{l}2x+\alpha=\frac{\pi}{2}-\alpha+k2\pi\\ 2x+\alpha=\pi-\frac{\pi}{2}+\alpha+k2\pi=a+\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-2\alpha+\frac{\pi}{2}+k2\pi\\ 2x=\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\alpha+\frac{\pi}{4}+k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)

4: \(2\cdot cos^2x-3\sqrt3\cdot\sin2x-4\cdot\sin^2x+4=0\)

=>\(2\cdot\frac{1+cos2x}{2}-3\sqrt3\cdot\sin2x-4\cdot\frac{1-cos2x}{2}+4=0\)

=>\(1+cos2x-3\sqrt3\cdot\sin2x-2\left(1-cos2x\right)+4=0\)

=>\(1+cos2x-3\sqrt3\cdot\sin2x-2+2\cdot cos2x+4=0\)

=>\(-3\sqrt3\cdot\sin2x+3\cdot cos2x+3=0\)

=>\(-\sqrt3\cdot\sin2x+cos2x=-1\)

=>\(\sqrt3\cdot\sin2x-cos2x=1\)

=>\(\sin2x\cdot\frac{\sqrt3}{2}-cos2x\cdot\frac12=\frac12\)

=>\(\sin\left(2x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 2x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{2}+k\pi\end{array}\right.\)

3: \(4\sqrt3\cdot\sin x\cdot cosx+4\cdot cos^2x-2\cdot\sin^2x=\frac52\)

=>\(2\sqrt3\cdot\sin2x+4\cdot\frac{1+cos2x}{2}-2\cdot\frac{1-cos2x}{2}=\frac52\)

=>\(2\sqrt3\cdot\sin2x+2+2\cdot cos2x-1+cos2x=\frac52\)

=>\(2\sqrt3\cdot\sin2x+3\cdot cos2x=\frac52-1=\frac32\)

=>\(\sin2x\cdot2+cos2x\cdot\sqrt3=\frac{\sqrt3}{2}\)

=>\(\sin2x\cdot\frac{2}{\sqrt7}+cos2x\cdot\frac{\sqrt3}{\sqrt7}=\frac{\sqrt3}{2\sqrt7}\)

=>\(\sin\left(2x+\alpha\right)=\frac{\sqrt3}{2\sqrt7}=\frac{\sqrt{21}}{14}\)

=>\(\left[\begin{array}{l}2x+\alpha=\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi\\ 2x+\alpha=\pi-\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi-\alpha\\ 2x=\pi-\arcsin\left(\frac{\sqrt{21}}{14}\right)-\alpha+k\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac12\cdot\arcsin\left(\frac{\sqrt{21}}{14}\right)+\frac{k\pi}{2}-\frac{\alpha}{2}\\ x=\frac{\pi}{2}-\frac12\cdot\arcsin\left(\frac{\sqrt{21}}{14}\right)-\frac{\alpha}{2}+\frac{k\pi}{2}\end{array}\right.\)

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