Chứng minh:
\(\sqrt{10+\sqrt{19}}.\sqrt{10-\sqrt{19}}\) = 9
Tính (Rút gọn):
a) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
b)\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
c)\(\left(\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}\right):2\sqrt{\sqrt{5}-2}\)
d)\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}\)
a)
\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(4^2-15)=2\)
b)
\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
c)
\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)
\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)
d)
\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)
\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)
Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)
\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)
\(\sqrt{10-\sqrt{19}}-\sqrt{10+\sqrt{19}}\)
Tính:
a,y=2\(+\sqrt{17-4\sqrt{9}+4\sqrt{5}}\)
b,t=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
c,x=\(\sqrt{19+8\sqrt{3}}+\sqrt{19-8\sqrt{3}}\)
b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))
t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)
t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))
t = (\(\sqrt{5}\) -1)2.(3 +\(\sqrt{5}\))
t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))
t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)
t = 6
cho \(\sqrt{x^2-6x+19}\)-\(\sqrt{x^2-6x+10}\)=3 . tính giá trị của T=\(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\)
\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)
\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)
\(\Rightarrow T=3\)
a,\(\sqrt{8+2\sqrt{15}}\) -\(\sqrt{6+2\sqrt{15}}\)
b, \(\sqrt{17-2\sqrt{72}}-\sqrt{19+2\sqrt{18}}\)
c, \(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\)
d, \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
e, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
Bài 5: So sánh
1,A=\(\sqrt{13}\) + \(\sqrt{20}\)
B=\(\sqrt{24}\) + \(\sqrt{19}\)
2,A=\(\sqrt{26}\) + \(\sqrt{10}\)
B=\(\sqrt{64}\)
Bài 2:
\(A=\sqrt{26}+\sqrt{10}>\sqrt{25}+\sqrt{9}=5+3=8\)
\(B=\sqrt{64}=8\)
Do đó: A>B
1.Ta có:
\(A=\)\(\sqrt{13}+\sqrt{20}=\sqrt{13}+2\sqrt{5}\)
\(B=\)\(\sqrt{24}+\sqrt{19}=\sqrt{19}+2\sqrt{6}\)
So sánh ta thấy:
\(\sqrt{13}<\sqrt{19}\) ; \(2\sqrt{5}<2\sqrt{6}\)
Vậy A < B
Cho \(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}=3\)
Tính M = \(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\)
Bài 31 (trang 19 SGK Toán 9 Tập 1)
a) So sánh $\sqrt{25-16}$ và $\sqrt{25}-\sqrt{16}$ ;
b) Chứng minh rằng, với $a>b>0$ thì $\sqrt{a}-\sqrt{b}<\sqrt{a-b}$.
a, Ta có \(\sqrt{25-16}=\sqrt{9}=3\)
\(\sqrt{25}-\sqrt{16}=5-4=1\)
Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
a) căn 25 - 16 > căn 25 - căn 16
b)Với nên đều xác định
Để so sánh và ta quy về so sánh và .
+) .
+)
.
Do nên
Do
(đpcm)
Vậy .
a) +) .
+) .
Vì nên .
Vậy .
b) Với nên đều xác định.
Để so sánh và ta quy về so sánh và .
+) .
+) .
Do nên
Do
(đpcm)
Vậy .
cho biểu thức A=\(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+19}=3\)
hãy tính giá trị của biểu thức A=\(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\)