\(VT=\frac{\left(5\sqrt{3}+5\sqrt{2}\right).\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)

\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)^2.\left(5-2\sqrt{6}\right)}{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5\sqrt{3}-5\sqrt{2}\right)}\)\(=\frac{\left(75+50\sqrt{6}+50\right).\left(5-2\sqrt{6}\right)}{75-50}\)

\(=\frac{25\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}{25}=5^2-\left(2\sqrt{6}\right)^2\)\(=25-24=1=VP\)

bn chép lại đề nhé

\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)

\(=\frac{\left(75+50\sqrt{6}+50\right)\left(\sqrt{3}-\sqrt{2}\right)}{75-50}\)

a chết bn ơi bài ở trên chưa đúng đâu, bài giải đây nè

\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)

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

\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=3-2=1\left(đpcm\right)\)

\(VT=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(a-b\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a^2-a\sqrt{ab}-b^2-b\sqrt{ab}-a^2+b^2}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{-\left(a+b\right)\sqrt{ab}}=\sqrt{b}-\sqrt{a}=VP\)

Vậy đẳng thức được chứng minh

Cảm ơn cậu nhiều nha ^^

\(A=\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\\ =\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\\ =\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{3}-\sqrt{2}}\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =3-2\\ =1\)

Vậy \(A=1\)

Bài 1:

a/ \(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}\)

\(=\frac{10}{2}=5\)

b/ \(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)

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

\(=\left(2-\sqrt{3}\right)\sqrt{2+4\sqrt{6}}\)

Bạn coi lại đề, tới đây ko rút gọn được nữa nên chắc bạn ghi đề nhầm ở chỗ nào đó

c/ \(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(5-\sqrt{24}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-\sqrt{24}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\left(5+2\sqrt{6}\right)\left(5-\sqrt{24}\right)=\left(5+\sqrt{24}\right)\left(5-\sqrt{24}\right)=1\)

d/ Nhân cả tử và mẫu của từng phân số với liên hợp của mẫu, mẫu số sẽ thành 1 hết:

\(=\frac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\frac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-1\)

\(=\sqrt{25}-1=5-1=4\)

Câu 2:

a/ ĐKXĐ: \(x\ge1\)

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

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

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=2\)

TH1: \(\sqrt{x-1}-1\ge0\Rightarrow x\ge2\) pt trở thành:

\(\sqrt{x-1}-1=2\Leftrightarrow\sqrt{x-1}=3\)

\(\Rightarrow x-1=9\Rightarrow x=10\) (nhận)

TH2: \(\sqrt{x-1}-1< 0\Rightarrow1\le x< 2\) pt trở thành:

\(1-\sqrt{x-1}=2\Rightarrow\sqrt{x-1}=-1< 0\) (vô nghiệm)

b/

\(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)

Do \(\left\{{}\begin{matrix}\sqrt{x^2-x}\ge0\\\sqrt{x^2+x-2}\ge0\end{matrix}\right.\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x^2-x}=0\\\sqrt{x^2+x-2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x^2+x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=1\)

\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(2-2\sqrt{2}.\sqrt{3}+3\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=3-2=1\)

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

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

\(=\frac{-2.2\sqrt{x}}{2}\)

\(=-2\sqrt{x}\)

Thank bạn bài vừa rồi đã k cho mk^^