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Câu 1/
\(\left\{{}\begin{matrix}\sqrt{\dfrac{4x}{5y}}=\sqrt{x+y}-\sqrt{x-y}\left(1\right)\\\sqrt{\dfrac{5y}{x}}=\sqrt{x+y}+\sqrt{x-y}\left(2\right)\end{matrix}\right.\)
Lấy (1).(2) vế theo vế được
\(\left(\sqrt{x+y}-\sqrt{x-y}\right)\left(\sqrt{x+y}+\sqrt{x-y}\right)=2\)
\(\Leftrightarrow x+y-\left(x-y\right)=2\)
\(\Leftrightarrow2y=2\)
\(\Leftrightarrow y=1\)
Thế vô tìm được x.
Câu 2/ Đề chưa đủ. x, y, z thuộc R luôn à. Tìm min hay max hay là tìm cả 2.
Ta có \(x+y+z=1\Rightarrow x+y=1-z,\) ta có:
\(\frac{x+y}{\sqrt{xy+z}}=\frac{1-z}{\sqrt{xy+1-x-y}}=\frac{1-z}{\sqrt{\left(1-x\right)\left(1-y\right)}}\)
\(\frac{y+z}{\sqrt{yz+x}}=\frac{1-x}{\sqrt{yz+1-y-z}}=\frac{1-x}{\sqrt{\left(1-y\right)\left(1-z\right)}}\)
\(\frac{z+x}{\sqrt{zx+y}}=\frac{1-y}{\sqrt{zx+1-x-z}}=\frac{1-y}{\sqrt{\left(1-x\right)\left(1-z\right)}}\)
Khi đó \(P=\frac{x+y}{\sqrt{xy+z}}+\frac{y+z}{\sqrt{yz+x}}+\frac{z+x}{\sqrt{zx+y}}=\frac{1-z}{\sqrt{\left(1-x\right)\left(1-y\right)}}+\frac{1-x}{\sqrt{\left(1-y\right)\left(1-z\right)}}+\frac{1-y}{\sqrt{\left(1-x\right)\left(1-z\right)}}\)
\(\ge3\sqrt[3]{\frac{1-z}{\left(1-x\right)\left(1-y\right)}\times\frac{1-x}{\left(1-y\right)\left(1-z\right)}\times\frac{1-y}{\left(1-x\right)\left(1-z\right)}}=3\)
Vậy \(MinP=3\) đạt được khi \(x=y=z=\frac{1}{3}\)
\(P=\dfrac{x+y}{\sqrt{xy+z}}+\dfrac{y+z}{\sqrt{yz+x}}+\dfrac{z+x}{\sqrt{xz+y}}\)
\(P=\dfrac{x+y}{\sqrt{xy+\left(x+y+z\right)z}}+\dfrac{y+z}{\sqrt{yz+\left(x+y+z\right)x}}+\dfrac{x+z}{\sqrt{zx+\left(x+y+z\right)y}}\)
\(P=\dfrac{x+y}{\sqrt{xy+xz+yz+z^2}}+\dfrac{y+z}{\sqrt{yz+x^2+xy+xz}}+\dfrac{x+z}{\sqrt{xz+xy+y^2+yz}}\)
\(P=\dfrac{x+y}{\sqrt{\left(x+z\right)\left(y+z\right)}}+\dfrac{y+z}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{x+z}{\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\sqrt{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}}}=3\sqrt[3]{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}=3\)
\(\Rightarrow P\ge3\)
Vậy \(P_{min}=3\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{1}{3}\)
a, Sửa đề \(xy=\dfrac{2}{7}\)
Ta có: \(xy=\dfrac{2}{7};yz=\dfrac{3}{2};zx=\dfrac{3}{7}\Rightarrow xy.yz.zx=\dfrac{2}{7}.\dfrac{3}{2}.\dfrac{3}{7}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{9}{49}\Leftrightarrow\left(xyz\right)^2=\left(\pm\dfrac{3}{7}\right)^2\Rightarrow\left[{}\begin{matrix}xyz=\dfrac{3}{7}\\xyz=-\dfrac{3}{7}\end{matrix}\right.\)
+) Xét trường hợp \(xyz=\dfrac{3}{7}\)\(\Rightarrow\dfrac{2}{7}.z=\dfrac{3}{7}\Rightarrow z=\dfrac{3}{7}:\dfrac{2}{7}=\dfrac{3}{2}\)
\(\Rightarrow y.\dfrac{3}{2}=\dfrac{3}{2}\Rightarrow y=1\Rightarrow x.1=\dfrac{2}{7}\Rightarrow x=\dfrac{2}{7}\)
+) Xét trường hợp \(xyz=-\dfrac{3}{7}\Rightarrow\dfrac{2}{7}.z=-\dfrac{3}{7}\Rightarrow z=-\dfrac{3}{2}\)
\(\Rightarrow y.\dfrac{-3}{2}=\dfrac{3}{2}\Rightarrow y=-1\Rightarrow x.\left(-1\right)=\dfrac{2}{7}\Rightarrow x=-\dfrac{2}{7}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=1\\z=\dfrac{2}{7}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-1\\z=-\dfrac{2}{7}\end{matrix}\right.\)
b, Ta có: \(xy=9z;yz=4x;zx=16y\Rightarrow\dfrac{xy}{z}=9;\dfrac{yz}{x}=4;\dfrac{zx}{y}=16\)
\(\Rightarrow\dfrac{xy}{z}.\dfrac{yz}{x}.\dfrac{zx}{y}=9.4.16\Rightarrow xyz=576\)
\(\Rightarrow xy=\dfrac{576}{z};yz=\dfrac{576}{x};zx=\dfrac{576}{y}\)
\(\Rightarrow\dfrac{576}{z}=9z\Rightarrow9z^2=576\Rightarrow z^2=64\Rightarrow z=\pm8\)
\(\dfrac{576}{x}=4x\Rightarrow4x^2=576\Rightarrow x^2=144\Rightarrow x=\pm12\)
\(\dfrac{576}{y}=16y\Rightarrow16y^2=576\Rightarrow y^2=36\Rightarrow y=\pm6\)
Vì xyz=156 => x;y;z dương hoặc trong x;y;z có 2 số âm
\(\Rightarrow\left\{{}\begin{matrix}x=12\\y=6\\z=8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=12\\y=-6\\z=-8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-12\\y=6\\z=-8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-12\\y=-6\\z=8\end{matrix}\right.\)
Vậy...
a) \(xy=\dfrac{3}{7};yz=\dfrac{3}{2};zx=\dfrac{3}{7}\)
từ \(xy=\dfrac{3}{7}vàzx=\dfrac{3}{7}\) \(\Rightarrow\) \(z=y\)
\(yz=\dfrac{3}{2}\) \(\Leftrightarrow\) \(y^2=\dfrac{3}{2}\) \(\Leftrightarrow\) \(y=\sqrt{\dfrac{3}{4}}\) \(\Leftrightarrow\) \(y=z=\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow\) \(xy=\dfrac{3}{7}\) \(\Leftrightarrow\) \(x.\dfrac{\sqrt{3}}{2}=\dfrac{3}{7}\) \(\Leftrightarrow\) \(x=\dfrac{3}{7}:\dfrac{\sqrt{3}}{2}\) = \(\dfrac{3}{7}.\dfrac{2}{\sqrt{3}}=\dfrac{6}{7\sqrt{3}}\) = \(\dfrac{2\sqrt{3}}{7}\)
vậy \(x=\dfrac{2\sqrt{3}}{7}\) ; \(y=\dfrac{\sqrt{3}}{2}\) ; \(z=\dfrac{\sqrt{3}}{2}\)
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
nhanh lên chiều nay tui nộp rùi
= \(\dfrac{\sqrt{xy}-1+\sqrt{yz}-3+\sqrt{zx}-5}{3+9+6}\) = \(\dfrac{11-\left(1+3+5\right)}{18}\)=\(\dfrac{1}{9}\)