\(\sqrt{xy}+1+\sqrt{x}+\sqrt{y}\)

\(=\left(\sqrt{xy}+\sqrt{y}\right)+\left(\sqrt{x}+1\right)\)

\(=\sqrt{y}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)

\(=\left(\sqrt{y}+1\right)\left(\sqrt{x}+1\right)\)

\(\sqrt{ab}+1+\sqrt{a}+\sqrt{b}\)

\(=(\sqrt{ab}+\sqrt{a})+\left(\sqrt{b}+1\right)\)

\(=\sqrt{a}\left(\sqrt{b}+1\right)+\left(\sqrt{b}+1\right)\)

\(=\left(\sqrt{b}+1\right)\left(\sqrt{a}+1\right)\)

\(\sqrt{8,1}.\sqrt{250}\)

\(=\sqrt{81}.\sqrt{25}\)

\(=9.5\)

\(=45\)

\(\sqrt{2,5}.\sqrt{360}\)

\(=\sqrt{25}.\sqrt{36}\)

\(=5.6\)

\(=30\)

\(\sqrt{\frac{-49}{-121}}=\sqrt{\frac{49}{121}}\)

\(=\frac{\sqrt{49}}{\sqrt{121}}\)

\(=\frac{7}{11}\)

\(\sqrt{\frac{-36}{-169}}=\sqrt{\frac{36}{169}}\)

\(=\frac{\sqrt{36}}{\sqrt{169}}=\frac{6}{13}\)

a, \(2+\sqrt{3x+4}=x\)(ĐKXĐ: \(x>\frac{3}{4}\))

\(\Leftrightarrow\sqrt{3x+4}=x-2\)

\(\Leftrightarrow\left(\sqrt{3x+4}\right)^2=\left(x-2\right)^2\)

\(\Leftrightarrow3x+4=x^2-4x+4\)

\(\Leftrightarrow x^2-4x+4-3x-4=0\)

\(\Leftrightarrow x^2-7x=0\)

\(\Leftrightarrow x\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(L\right)\\x=7\left(TM\right)\end{cases}}}\)

Vậy PT có nghiệm là \(x=7\)

b, \(\sqrt{4x^2-4x+1}-\sqrt{9x^2}=0\)

\(\Leftrightarrow\sqrt{4x^2-4x+1}=\sqrt{9x^2}\)

\(\Leftrightarrow\left(\sqrt{4x^2-4x+1}\right)^2=\left(\sqrt{9x^2}\right)^2\)

\(\Leftrightarrow4x^2-4x+1=9x^2\)

\(\Leftrightarrow9x^2-4x^2+4x-1=0\)

\(\Leftrightarrow5x^2+4x-1=0\)

\(\Leftrightarrow\left(x-\frac{1}{5}\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{5}=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)

Vậy PT có nghiệm là \(x\in\left\{-1;\frac{1}{5}\right\}\)

A B C H I K

a)

Ta có:

Tam giác AKC vuông tại K \(\Rightarrow sinA=\frac{KC}{AC}\)

\(VT=S_{ABC}=\frac{1}{2}.AB.CK=\frac{1}{2}.AB.\left(AC.\frac{KC}{AC}\right)=\frac{1}{2}.AB.AC.sinA=VP\)(đpcm)

b)

\(\left(1-cos^2A-cos^2B-cos^2C\right).S_{ABC}\)

\(=\left(1-\frac{KC^2}{AC^2}-\frac{BI^2}{AB^2}-\frac{AH^2}{BC^2}\right).S_{ABC}\)

\(=\left[\left(1-\frac{AH^2}{BC^2}\right)-\left(\frac{KC^2}{AC^2}+\frac{BI^2}{AB^2}\right)\right].S_{ABC}\)

\(=\left(\left(1-\frac{AH^2}{BC^2}\right)-\frac{AB^2.KC^2-AC^2.BI^2}{AB^2.AC^2}\right).S_{ABC}\)

\(=\left(\left(1-\frac{AH^2}{BC^2}\right)-\frac{S^2_{ABC}-S^2_{ABC}}{AB^2.AC^2}\right).S_{ABC}\)

\(=\left(1-\frac{AH^2}{BC^2}\right).S_{ABC}=S_{ABC}-\frac{AH^2}{BC^2}.S_{ABC}\)

\(\hept{\begin{cases}\frac{x^2}{y}+x=2\\\frac{y^2}{x}+y=\frac{1}{2}\end{cases}}\) ĐK: x,y khác 0

Nhân theo vế hai pt ta được

\(\left(\frac{x^2}{y}+x\right).\left(\frac{y^2}{x}+y\right)=2\)

\(\Leftrightarrow xy+x^2+y^2+xy=1\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+y=1\\x+y=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1-y\\x=-1-y\end{cases}}}\)

Xét 2 trường hợp trên bằng cách thế x theo y là được

sao nhân theo 2 vế lại bằng 2

=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}^2+\sqrt{4-\sqrt{10+2\sqrt{5}}}^2\)

=\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}\)

=\(8\)

\(A=\sqrt{4+\sqrt{10+\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(>0\)

=>  \(A^2=4+\sqrt{10+2\sqrt{5}}+2.\sqrt{\left(4-\sqrt{10+2\sqrt{5}}\right)\left(4+\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10-2\sqrt{5}}\)

\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

\(=8+2\sqrt{6-2\sqrt{5}}\)

\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=8+2\left(\sqrt{5}-1\right)\)

\(=6+2\sqrt{5}\)

=>  \(A=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

\(ĐK:x>-8\)

Nhân cả 2 vế của pt với \(\sqrt{x+8}\)

\(PT\Leftrightarrow\left(x+8\right)+9x-6\sqrt{x}.\sqrt{x+8}=0\)

\(\Leftrightarrow\left(x+8\right)-2\sqrt{9x}.\sqrt{x+8}+9x=0\)

\(\Leftrightarrow\left(\sqrt{x+8}-3x\right)^2=0\)

\(\Leftrightarrow\sqrt{x+8}-3x=0\)

\(\Leftrightarrow\sqrt{x+8}=3x\)

\(\Rightarrow\hept{\begin{cases}x\ge0\\x+8=9x^2\end{cases}\Rightarrow x=1}\)

Vậy pt có nghiệm x=1