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1/ \(\Leftrightarrow\left\{{}\begin{matrix}2x^2-4xy+2x-4y+6=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2-2xy+4x-4y+4=0\)
\(\Leftrightarrow\left(x-y\right)^2+4\left(x-y\right)+4=0\)
\(\Leftrightarrow\left(x-y+2\right)^2=0\)
\(\Rightarrow y=x+2\)
Thay vào 1 trong 2 pt ban đầu là xong
2/ \(x^2-\left(y+2\right)x-6y^2+11y-3=0\)
\(\Delta=\left(y+2\right)^2-4\left(-6y^2+11y-3\right)\)
\(=25y^2-40y+16=\left(5y-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{y+2+5y-4}{2}\\x=\frac{y+2-5y+4}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3y-1\\x=-2y+3\end{matrix}\right.\)
Thay vào pt 2 là được
c/ \(S=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)
\(S< 1+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)
\(S< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)
\(S< 1+2\left(\sqrt{100}-1\right)=19\)
\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{101}-\sqrt{100}}\)
\(S>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)
\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)
\(\Rightarrow18< S< 19\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải số tự nhiên
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=4\\\frac{1}{\left(x+1\right)^2-1}+\frac{1}{\left(y+1\right)^2-1}=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\frac{1}{\left(x+1\right)^2-1}+\frac{1}{\left(y+1\right)^2-1}=\frac{2}{3}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\y+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=4\\\frac{1}{a^2-1}+\frac{1}{b^2-1}=\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{a^2-1}+\frac{1}{\frac{16}{a^2}-1}=\frac{2}{3}\)
\(\Rightarrow a^4-8a^2+16=0\Rightarrow a^2=4\Rightarrow a=\pm2\Rightarrow x=...\)
b/ ĐKXĐ: ...
\(\Rightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)
\(\Rightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)
\(\Rightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\sqrt{2-\frac{1}{y}}+xy\sqrt{2-\frac{1}{x}}}=0\)
\(\Rightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\sqrt{2-\frac{1}{y}}+xy\sqrt{2-\frac{1}{x}}}=0\right)\)
\(\Rightarrow\sqrt{y}=\sqrt{x}\Rightarrow y=x\) (ngoặc phía sau luôn dương)
Thay vào pt đầu:
\(\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}=2\)
Mặt khác áp dụng BĐT \(a+b\le\sqrt{2\left(a^2+b^2\right)}\)
\(\Rightarrow\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}\le\sqrt{2\left(\frac{1}{x}+2-\frac{1}{x}\right)}=2\)
Dấu "=" xảy ra khi và chỉ khi:
\(\frac{1}{\sqrt{x}}=\sqrt{2-\frac{1}{x}}\Rightarrow\frac{1}{x}=2-\frac{1}{x}\Rightarrow x=1\Rightarrow y=1\)
Ai phát hiện sai đề thì sửa và làm giúp mk hộ với, cảm ơn :) (chỉ cần làm tóm tắt thôi)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
ĐKXĐ x ; y > 0
(1) \(\Rightarrow\left(y-x\right)\left(\frac{1}{\sqrt{x}y}+x+2xy\right)=0\)
\(\Rightarrow x=y\)
\(\Rightarrow...\)
#Kaito#
\(P=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\sqrt{2}\left(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\right)\)
\(=\sqrt{2}\left(\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\right)=\sqrt{2}\)
2/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{y^2-\frac{7}{y^2}}=a\\\sqrt{y-\frac{7}{y^2}}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a^2-b^2=y^2-y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=y\\\left(a-b\right)\left(a+b\right)=y^2-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a-b=y-1\end{matrix}\right.\)
\(\Rightarrow b=\frac{1}{2}\Rightarrow\sqrt{y-\frac{7}{y^2}}=\frac{1}{2}\Rightarrow y-\frac{7}{y^2}=\frac{1}{4}\Rightarrow4y^3-y^2-28=0\)
\(\Rightarrow y=2\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)
\(\Rightarrow3x^2-xy-2y^2=0\)
\(\Rightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-\frac{2}{3}y\end{matrix}\right.\) thay vào 1 trong 2 pt là xong