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$\sin18=\cos72=2 \cos^{2}36-1=2(1- \sin^{2}18)^{2}-1
\Leftrightarrow 8 \sin^{4}18 -8 \sin^{2}18- \sin18+1=0
\Leftrightarrow ( \sin18-1)[8 \sin^{3}18+8 \sin^{2}18-1]=0 $
ht
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Ta có :
4 = 22
6 = 2 . 3
=> ƯCLN ( 4 ; 6 ) = 2
Vậy ƯCNN ( 4 ; 6 ) = 2
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\(1,\left(\frac{\left(x+1\right)^2.\left(y+1\right)^2}{\left(x+1\right)^2}+\frac{\left(x+1\right)^2\left(y+1\right)^2}{\left(y+1\right)^2}\right)\left(xy+1\right)\ge\left(x+1\right)^2\left(y+1\right)^2\)
\(\left[\left(y+1\right)^2+\left(x+1\right)^2\right]\left(xy+1\right)\ge\left(xy+y+x+1\right)^2\)
\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)\ge\left(xy+y+x+1\right)^2\)
\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)-\left(xy+y+x+1\right)^2\ge0\)
\(\left(y^2+2y+1+x^2+2x+1\right)\left(xy+1\right)-\left(x^2+2x+1\right)\left(y^2+2y+1\right)\ge0\)
\(xy\left(x-1\right)^2+\left(xy-1\right)^2\ge0\)
\(< =>BĐT\)luôn đúng
dấu "=" xảy ra khi \(x=y=1\)
mình ko chắc đã đúngg đâu
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ĐK: \(x^2-3y^2+30\ge0\).
Phương trình thứ nhất tương đương với:
\(\left(x-y+3\right)\left(x+2y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y-3\\x=1-2y\end{cases}}\)
Với \(x=y-3\)thế vào phương trình thứ hai ta được:
\(\sqrt{\left(y-3\right)^2-3y^2+30}+y-3-2y-5=0\)
\(\Leftrightarrow\sqrt{-2y^2-6y+39}=y+8\)
\(\Rightarrow-2y^2-6y+39=y^2+16y+64\)
\(\Leftrightarrow\orbr{\begin{cases}y=\frac{-11+\sqrt{46}}{3}\Rightarrow x=\frac{-20+\sqrt{46}}{3}\\y=\frac{-11-\sqrt{46}}{3}\Rightarrow x=\frac{-20-\sqrt{46}}{3}\end{cases}}\)
Thử lại thỏa mãn.
Với \(x=1-2y\)làm tương tự, thu được thêm một nghiệm là: \(x=\frac{17-2\sqrt{61}}{5},y=\frac{-6+\sqrt{61}}{5}\).
1a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
2. a) ĐK: x \(\ge\)4
Ta có: \(\sqrt{16x-64}=2\) <=> \(4\sqrt{x-4}=2\) <=> \(\sqrt{x-4}=\frac{1}{2}\)
<=> \(x-4=\frac{1}{4}\) <=> \(x=\frac{17}{4}\left(tm\right)\)
b) \(16x^2-\left(1+\sqrt{2}\right)^2=0\) <=> \(\left(4x-1-\sqrt{2}\right)\left(4x+1+\sqrt{2}\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{1+\sqrt{2}}{4}\\x=\frac{-1-\sqrt{2}}{4}\end{cases}}\)
c)Đk: x \(\ge\)0
\(x-2\sqrt{3x}+3=4\) <=> \(\left(\sqrt{x}-\sqrt{3}\right)^2=4\)
<=> \(\left(\sqrt{x}-\sqrt{3}-2\right)\left(\sqrt{x}-\sqrt{3}+2\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+2\\\sqrt{x}=\sqrt{3}-2\left(loại\right)\end{cases}}\)
<=> \(x=5+4\sqrt{3}\)
3a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{20}.\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)
\(=4\sqrt{5}-\frac{8\left(1+\sqrt{5}\right)}{5-1}=4\sqrt{5}-2\left(1+\sqrt{5}\right)=4\sqrt{5}-2-2\sqrt{5}=2\sqrt{5}-2\)
b) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}+9\sqrt{2}}\)
\(=\frac{2\left(2\sqrt{2}-3\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{6}\left(\sqrt{5}+3\sqrt{3}\right)}=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)