a) \(\left(-\dfrac{1}{3}\sqrt{63}\right)^2=\dfrac{1}{9}\cdot63=7\)
\(\left(-2\sqrt{2}\right)^2=8\)
mà 7<8
nên \(-\dfrac{1}{3}\sqrt{63}>-2\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{55}\right)^2=4\cdot55=220\)
\(\left(\dfrac{3}{5}\sqrt{750}\right)=\dfrac{9}{25}\cdot750=270\)
mà 220<270
nên \(2\sqrt{55}< \dfrac{3}{5}\sqrt{750}\)
hay \(-2\sqrt{55}< -\dfrac{3}{5}\sqrt{750}\)
1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
Tính:
a, A= \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)+ \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
b, B= \(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)+ \(\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
c, C= \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
1 ) thực hiện phép tính
b) \(4\sqrt{\dfrac{9}{2}}+\sqrt{2}+\sqrt{\dfrac{1}{18}}\)
c) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
d) \(\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}-\dfrac{4}{3-\sqrt{5}}\)
e) \(\dfrac{\sqrt{7}+7}{\sqrt{7}+1}-\dfrac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\dfrac{2\sqrt{35}-2\sqrt{7}}{1-\sqrt{5}}\)
3) giải phương trình
a) \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\)
ai giúp e với ! làm câu nào cũng được hết ạ
Rút gọn :
\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}\)
Rút gọn
\(A=\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}+\dfrac{1}{\sqrt{2}-\sqrt{3}}+....+\dfrac{1}{\sqrt{n-1}-\sqrt{n}}\) (n thuộc N, n>=2)
1 Đúng hoặc Sai,nếu sai thì sửa lại cho đúng
a/\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\) ; b/\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{10}\) ; c/\(\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-1\) ; d/\(\dfrac{8}{2\sqrt{8}-1}=\dfrac{P\left(2\sqrt{8}+1\right)}{4P-1}\) ; e/\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
2 Rút gọn các biểu thức
a/\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\) ; b/\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\) ; c/\(\dfrac{3+\sqrt{3}}{3-\sqrt{3}}+\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\) ; d/\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}+\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}}\)
trục căn thức ở mẫu vs giả thiết các biểu thức chữ đều có nghĩa ( từ bài 50 đến bài 52)
BT50
\(\dfrac{5}{\sqrt{10}}\);\(\dfrac{5}{2\sqrt{5}}\);\(\dfrac{1}{3\sqrt{20}}\);\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}\);\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}\)
BT51
\(\dfrac{3}{\sqrt{3}+1}\);\(\dfrac{2}{\sqrt{3}-1}\);\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\);\(\dfrac{b}{3+\sqrt{b}}\);\(\dfrac{p}{2\sqrt{p}-1}\)
BT52
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}\);\(\dfrac{3}{\sqrt{10}+\sqrt{7}}\);\(\dfrac{1}{\sqrt{x}-\sqrt{y}}\);\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}\)
Tìm Min và Max(nếu có)
A=2x-\(\sqrt{x}\)
B=x+\(\sqrt{x}\)
C=1+\(\sqrt{2-x}\)
D=\(\sqrt{-x^2+2x+5}\)
E=\(\dfrac{1}{2x-\sqrt{x}+3}\)
F=\(\dfrac{1}{3-\sqrt{1-x^2}}\)
B1: thực hiện phép tính
a )\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)
b ) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c )\(\dfrac{\sqrt{3-\sqrt{5}.}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
d ) \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
B2:chúng minh vế phải bằng vế trái
a) \(\dfrac{21+8\sqrt{5}}{4+\sqrt{5}}.\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
b) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=-2\sqrt{3}\)
