a, Áp dụng định lí Pytago cho tam giác ABC vuông tại A

\(BC^2=AB^2+AC^2\Rightarrow AC^2=BC^2-AB^2=900-576=324\Rightarrow AC=18\)cm 

* Áp dụng hệ thức :

 \(AH.BC=AB.AC\Rightarrow AH=\frac{AB.AC}{BC}=\frac{24.18}{30}=\frac{72}{5}\)cm 

* Áp dụng hệ thức : \(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{576}{30}=\frac{96}{5}\)cm 

\(CH=BC-BH=30-\frac{96}{5}=\frac{54}{5}\)cm

Ta co \(\sqrt{2012}-\sqrt{2011}=\frac{\left(\sqrt{2012}-\sqrt{2011}\right)\left(\sqrt{2012}+\sqrt{2011}\right)}{\sqrt{2012}+\sqrt{2011}}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)

\(\sqrt{2013}-\sqrt{2012}=\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2013}+\sqrt{2012}}=\frac{1}{\sqrt{2013}+\sqrt{2012}}\)

Do \(\sqrt{2012}+\sqrt{2011}< \sqrt{2013}+\sqrt{2012}\)

Suy ra \(\frac{1}{\sqrt{2012}+\sqrt{2011}}>\frac{1}{\sqrt{2013}+\sqrt{2012}}\)hay \(\sqrt{2012}-\sqrt{2011}>\sqrt{2013}-\sqrt{2012}\)

\(\sqrt{2012}-\sqrt{2011}>\sqrt{2013}-\sqrt{2012}\)

\(5,A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(A=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(A=\left|2x-1\right|+\left|2x-3\right|\)

\(A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(A\ge2\)

\(< =>MIN:A=2\)dấu = xảy khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

thế cũng nói tui báo cáo á

\(\left(x+1\right)^2y+\left(y+1\right)^2x=1\)

\(\Leftrightarrow x^2y+xy^2+4xy+x+y+4=5\)

\(\Leftrightarrow\left(xy+1\right)\left(x+y+4\right)=5\)

Đến đây bạn xét bảng giá trị, thu được kết quả là: 

\(\left(x,y\right)\in\left\{\left(0,1\right),\left(1,0\right),\left(-6,1\right),\left(1,-6\right)\right\}\).

\(x\left(4-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\4-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\4-x\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x\le4\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\x\ge4\end{cases}}\)

\(\Leftrightarrow0\le x\le4\).

\(x=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)

\(=2\)

Với \(x=2\):

\(\frac{\sqrt{\frac{1}{x}+4+4x}}{\sqrt{x}\left(2x^2-x-1\right)}=\frac{\sqrt{\frac{1}{2}+4+8}}{\sqrt{2}\left(8-2-1\right)}=\frac{1}{2}\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

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