Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)
b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)
c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
Lời giải:
Thay $3=xy+yz+xz$ vào biểu thức:
\(P=\frac{x}{\sqrt{x^2+xy+yz+xz}}+\frac{y}{\sqrt{y^2+xy+yz+xz}}+\frac{z}{\sqrt{z^2+xy+yz+xz}}\)
hay \(P=\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT Cauchy ta có:
\(\frac{x}{\sqrt{(x+y)(x+z)}}\leq \frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Hoàn toàn tương tự:
\(\frac{y}{\sqrt{(y+z)(y+x)}}\leq \frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\)
\(\frac{z}{\sqrt{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{z}{z+y}+\frac{z}{x+z}\right)\)
Cộng theo vế:
\(\Rightarrow P\leq \frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Vậy \(P_{\max}=\frac{3}{2}\). Dấu bằng xảy ra khi \(x=y=z=1\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
a: \(=\dfrac{\left(1-\sqrt{2}\right)^2}{1-\sqrt{2}}=1-\sqrt{2}\)
b: \(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{x-y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
d: \(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x-y}=\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
Lời giải:
Câu a:
Áp dụng BĐT Cô-si ngược dấu ta có:
\(\sqrt{3(x-3)}\leq \frac{3+(x-3)}{2}=\frac{x}{2}\)
\(\Rightarrow \sqrt{x-3}\leq \frac{x}{2\sqrt{3}}\Rightarrow \frac{\sqrt{x-3}}{x}\leq \frac{1}{2\sqrt{3}}\)
Hoàn toàn tương tự: \(\frac{\sqrt{y-3}}{y}\leq \frac{1}{2\sqrt{3}}\)
\(\Rightarrow p=\frac{\sqrt{x-3}}{x}+\frac{\sqrt{y-3}}{y}\leq \frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}\)
Dấu "=" xảy ra khi \(3=x-3; 3=y-3\Rightarrow x=y=6\)
Vậy \(p_{\max}=\frac{\sqrt{3}}{3}\Leftrightarrow x=y=6\)
Câu b: Các phân thức của $q$ là nghịch đảo của $p$ nên $q$ có min thôi em nhé. Nếu tìm min thì tương tự như câu a.
dạ