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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: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)
\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)
\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)
b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)
c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)
\(x=\dfrac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{\left(\sqrt[3]{2}+1\right)\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{3}=3\sqrt[3]{2}+1\)
\(y=\dfrac{6}{\sqrt[3]{16}+\sqrt[3]{4}+4}=\dfrac{6}{\sqrt[3]{4}\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}.3}=\dfrac{2\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}}=\dfrac{2\sqrt[3]{4}}{\sqrt[3]{4}}-\dfrac{\sqrt[3]{8}}{\sqrt[3]{4}}=2-\sqrt[3]{2}\)
=> x + y = \(\sqrt[3]{2}+1+2-\sqrt[3]{2}=3\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
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: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)