Rút câu dễ nhất :))

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

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

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

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

\(=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)

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

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

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

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

\(=>K=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(=>T=\sqrt{5}+1-\sqrt{5}=1\)

Bt làm câu 2 nhưng nhác đánh máy wa , còn câu 3 thì bó tay

thánh nào giúp tui CÂU 3 với Nguyễn Huy Tú

Toshiro Kiyoshisoyeon_Tiểubàng giảiAkai HarumaNguyễn Huy ThắngPhương AnÁi Hân NgôNguyễn Thanh HằngHung nguyenFairy TailĐời về cơ bản là buồn... cười!!!Linh_Windy,...

trần hữu tuyển

Lời giải:

Ta có:
\(A-B=(\sqrt{2016}-\sqrt{2014})+(\sqrt{2017}-\sqrt{2015})+(\sqrt{2018}-\sqrt{2022})\)

\(=\frac{2}{\sqrt{2016}+\sqrt{2014}}+\frac{2}{\sqrt{2017}+\sqrt{2015}}-\frac{4}{\sqrt{2018}+\sqrt{2022}}\)

Dễ thấy:

\(0< \sqrt{2016}+\sqrt{2014}< \sqrt{2018}+\sqrt{2022}; 0< \sqrt{2017}+\sqrt{2015}< \sqrt{2018}+\sqrt{2022}\)

\(\Rightarrow \frac{1}{\sqrt{2016}+\sqrt{2014}}>\frac{1}{\sqrt{2018}+\sqrt{2022}};\frac{1}{\sqrt{2017}+\sqrt{2015}}>\frac{1}{\sqrt{2018}+\sqrt{2022}}\)

\(\Rightarrow A-B=2\left(\frac{1}{\sqrt{2016}+\sqrt{2014}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}+\frac{1}{\sqrt{2017}+\sqrt{2015}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}\right)>0\)

\(\Rightarrow A>B\)

Lời giải:

Ta có:
\(A-B=(\sqrt{2016}-\sqrt{2014})+(\sqrt{2017}-\sqrt{2015})+(\sqrt{2018}-\sqrt{2022})\)

\(=\frac{2}{\sqrt{2016}+\sqrt{2014}}+\frac{2}{\sqrt{2017}+\sqrt{2015}}-\frac{4}{\sqrt{2018}+\sqrt{2022}}\)

Dễ thấy:

\(0< \sqrt{2016}+\sqrt{2014}< \sqrt{2018}+\sqrt{2022}; 0< \sqrt{2017}+\sqrt{2015}< \sqrt{2018}+\sqrt{2022}\)

\(\Rightarrow \frac{1}{\sqrt{2016}+\sqrt{2014}}>\frac{1}{\sqrt{2018}+\sqrt{2022}};\frac{1}{\sqrt{2017}+\sqrt{2015}}>\frac{1}{\sqrt{2018}+\sqrt{2022}}\)

\(\Rightarrow A-B=2\left(\frac{1}{\sqrt{2016}+\sqrt{2014}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}+\frac{1}{\sqrt{2017}+\sqrt{2015}}-\frac{1}{\sqrt{2018}+\sqrt{2022}}\right)>0\)

\(\Rightarrow A>B\)

giúp vs tth Trần Thanh Phương Nguyễn Văn Đạt Nguyễn Việt Lâm Akai Haruma

a.

\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)

c.

\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)

d.

\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)

Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì

\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)

Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)

Vậy \(\left(a;b\right)=\left(9;4\right)\)

1/ Tính: \(A=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{10}+1\right)^2}}{2\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}+\sqrt{\left(2\sqrt{2}+2\right)^2}}=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}=\dfrac{2\sqrt{2}-1}{6\sqrt{2}-3}=\dfrac{2\sqrt{2}-1}{3\left(2\sqrt{2}-1\right)}=\dfrac{1}{3}\)

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

Bài này ko khó,quá khó là đằng khác

hazzzzzz đăng lên đây thầy cô cũng ko giải , ko thành viên nào giải chỉ toàn thấy cmr tào lao, thui đi kiếm trang khác hỏi

Bai này nếu bạn ko giải dc thì cũng có số it người giải dc. sao lại than phiền?

Nên đây để học hỏi mà.( ko có người làm dc cũng là bt (trong dó có bạn)