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\(P=\dfrac{\left(\sqrt{a+1}+1\right)\left(\sqrt{a+1}+2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}+\dfrac{2\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}}{a-3}+\dfrac{2a+2-4\sqrt{a+1}}{a-3}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}+2a+2-4\sqrt{a+1}-2-5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{3a+3-6\sqrt{a+1}}{a-3}\)
Có thể dừng ở đây hoặc nếu thích thì làm tiếp như sau (chưa chắc gọn hơn):
\(P=\dfrac{3\left(a+1\right)-6\sqrt{a+1}}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}=\dfrac{3\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}\)
\(P=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}-2}\)
Cho \(5\sqrt{x}7\) mk viet nham
Sua lai thanh \(5\sqrt{x}-7\)
a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)
=>10 căn x+5-5 chia hết cho 2 căn x+1
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
hay \(x\in\varnothing\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
♡
\(\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(=\dfrac{2\left(1+\sqrt{2}\right)-2\left(1-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)
\(=\dfrac{2+2\sqrt{2}-2+2\sqrt{2}}{1-2}=-4\sqrt{2}\)
♡
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left[-\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-3\)
♡
\(\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(=\dfrac{2\left(7-4\sqrt{3}\right)+2\left(7+4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{49-48}\)
= 28
♡
\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{4}{6-2\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)-2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\sqrt{5}-2-2\sqrt{5}-2}{5-1}\)
= - 1
♡
\(\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(=\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)}\)
\(=-2-2\sqrt{3}-\sqrt{3}=-2-3\sqrt{3}\)
♡
\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\) (nhân [căn 2] vào cả tử và mẫu)
\(=\dfrac{2}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{2\left(5-\sqrt{5}\right)}{25-5}=\dfrac{5-\sqrt{5}}{10}\)
1/
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{4-3}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
2/
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(5-2\right)=-3\)
#F.C
b: \(B=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{5}+\dfrac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
1: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
2: \(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}=\dfrac{8}{2}=4\)
4: \(=\dfrac{-3+5\sqrt{3}}{11}+\dfrac{3+5\sqrt{3}}{11}=\dfrac{10\sqrt{3}}{11}\)
\(P=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}+\sqrt{3+\sqrt{5}}\right)}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{4+\sqrt{6+2\sqrt{5}}}+\dfrac{3\sqrt{2}-\sqrt{10}}{4-\sqrt{6-2\sqrt{5}}}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{5+\sqrt{5}}+\dfrac{3\sqrt{2}-\sqrt{10}}{5-\sqrt{5}}\)
\(=\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(5-\sqrt{5}\right)+\left(3\sqrt{2}-\sqrt{10}\right)\left(5+\sqrt{5}\right)}{20}\)
\(=\dfrac{15\sqrt{2}-3\sqrt{10}+5\sqrt{10}-5\sqrt{2}+15\sqrt{2}+3\sqrt{10}-5\sqrt{10}-5\sqrt{2}}{20}\)
\(=\dfrac{30\sqrt{2}-10\sqrt{2}}{20}=\dfrac{20\sqrt{2}}{20}=\sqrt{2}\)
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