Thực hiện phép tính:\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\) -\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
78, Với \(x>0;x\ne1\)
\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{x-1}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\frac{x-1}{\sqrt{x}}\)
79, Với a > 0 ; \(a\ne1\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(=\left(\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(=\left(\frac{4\sqrt{a}+4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)=\frac{4a\sqrt{a}\left(a+1\right)}{\sqrt{a}\left(a-1\right)}=\frac{4a\left(a+1\right)}{a-1}\)
a, Theo định lí Pytago tam giác ABH : \(AB^2=AH^2+BH^2\)(1)
Theo định lí Pytago tam giác ACH : \(AC^2=AH^2+AC^2\)(2)
Lấy (1) - (2) : \(AB^2-AC^2=AH^2+BH^2-AH^2-HC^2\)
\(\Leftrightarrow AB^2-AC^2=BH^2-HC^2\Leftrightarrow AB^2+HC^2=BH^2+AC^2\)
b, Ta có : \(AH^2=AM.AB\)( hệ thức lượng ) (1)
\(AH^2=AN.AC\)( hệ thức lượng ) (2)
Từ (1) ; (2) suy ra : \(AM.AB=AN.AC\)(3)
(3) => \(\frac{AM}{AC}=\frac{AN}{AB}\)
Xét tam giác AMN và tam giác ACB ta có :
^A _ chung
\(\frac{AM}{AC}=\frac{AN}{AB}\)
Vậy tam giác AMN ~ tam giác ACB ( c.g.c )
\(M=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x-1}}{1}\)
\(=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(M< \frac{3}{2}\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}+1}< \frac{3}{2}\Leftrightarrow2\left(2\sqrt{x}+1\right)< 3\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow4\sqrt{x}+2< 3\sqrt{x}+3\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\).
\(\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\Rightarrow0\le x< 9\)
anh đi anh nhớ quê nha
nhớ canh rau muống nhớ cà dầm tương
nhớ thằng đẩy bố xuống mương
bố mà bắt được bố tương vỡ mồm
\(A=\sqrt{6+3\sqrt{3}}-\sqrt{14-3\sqrt{3}}-2\sqrt{2}\)
\(A\sqrt{2}=\sqrt{12+6\sqrt{3}}-\sqrt{28-6\sqrt{3}}-4\)
\(=\sqrt{9+2.3.\sqrt{3}+3}-\sqrt{27-2.3\sqrt{3}.1+1}-4\)
\(=\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{3}-1\right)^2}-4\)
\(=3+\sqrt{3}-3\sqrt{3}+1-4\)
\(=-2\sqrt{3}\)
Suy ra \(A=-\sqrt{6}\).
ta có
\(A=\frac{\sqrt{x}-\sqrt{x-1}-\left(\sqrt{x}+\sqrt{x-1}\right)}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{0}{1-\sqrt{x}}\)
\(=-\frac{2\sqrt{x-1}}{x-\left(x-1\right)}=-2\sqrt{x-1}\) dễ thấy \(A\le0\) với mọi x
a) \(A=\frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
b) \(B=\left(\frac{\sqrt{x}-\sqrt{y}\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{x+y+2\sqrt{xy}}{x-y}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)\(=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1\)
\(ĐKXĐ:x\ge0;x\ne1\)
\(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\)
\(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+1\right)}=\frac{\sqrt{x}-1}{x+1}\)
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}-\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}-2\sqrt{16-\left(10+2\sqrt{5}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8-2\sqrt{6-2\sqrt{5}}=8-2\sqrt{\left(\sqrt{5}-1\right)^2}=8-2\left(\sqrt{5}-1\right)\)
\(=8-2\sqrt{5}+2=10-2\sqrt{5}\)
Vậy \(A=\sqrt{10-2\sqrt{5}}\)