Bài 4: Liên hệ giữa phép chia và phép khai phương
Các câu hỏi tương tự
B1: Tính:a, sqrt{72}divsqrt{8}b, (sqrt{28}-sqrt{7}+sqrt{112})divsqrt{7}B2: Tính:a, sqrt{dfrac{49}{8}}divsqrt{3dfrac{1}{8}}b, sqrt{54x}divsqrt{6x}c, sqrt{dfrac{1}{125}}timessqrt{dfrac{32}{35}}divsqrt{dfrac{56}{225}}giúp em với ạ , em cảm mơn
Đọc tiếp
B1: Tính:
a, \(\sqrt{72}\div\sqrt{8}\)
b, \((\sqrt{28}-\sqrt{7}+\sqrt{112})\div\sqrt{7}\)
B2: Tính:
a, \(\sqrt{\dfrac{49}{8}}\div\sqrt{3\dfrac{1}{8}}\)
b, \(\sqrt{54x}\div\sqrt{6x}\)
c, \(\sqrt{\dfrac{1}{125}}\times\sqrt{\dfrac{32}{35}}\div\sqrt{\dfrac{56}{225}}\)
giúp em với ạ , em cảm mơn 
Rút gọn biểu thức
E = \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}-4}{4-x}\)
tính:a) sqrt{dfrac{1}{8}}.sqrt{2}.sqrt{125}.sqrt{dfrac{1}{5}}b)sqrt{sqrt{2}-1}.sqrt{sqrt{2}+1}c) sqrt{11-6sqrt{2}}.sqrt{11+6sqrt{2}}d) sqrt{12-6sqrt{3}}.sqrt{dfrac{1}{3-sqrt{3}}}e) dfrac{sqrt{15}-sqrt{6}}{sqrt{35}-sqrt{14}}f) dfrac{2sqrt{15}-2sqrt{10}+sqrt{6}-3}{2sqrt{5}-2sqrt{10}-sqrt{3}+sqrt{6}}g) left(dfrac{1}{5-2sqrt{6}}+dfrac{2}{5+2sqrt{6}}right)left(15+2sqrt{6}right)
Đọc tiếp
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
16 tháng 5 2023 lúc 21:06
Trục căn thức ở mẫu:
B = \(\dfrac{1+\sqrt{5}}{2-\sqrt{5}}\) C = \(\dfrac{5-\sqrt{x}}{2\sqrt{x}}\)
D = \(\dfrac{\sqrt{a}+1}{2\sqrt{a}-1}\) E = \(\dfrac{15}{5\sqrt{3}-3\sqrt{5}}\)
\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4\cdot2-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4}\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\)
1. Tìm x để bt có nghĩa
Adfrac{sqrt{2x+3}}{sqrt{x-3}}
Bsqrt{dfrac{2x+3}{x-3}}
Csqrt{-dfrac{5}{x+2}}
Dsqrt{-x}+dfrac{1}{x+3}
2. Rút gọn bt
Asqrt{dfrac{a+sqrt{a^2-1}}{2}}-sqrt{dfrac{a-sqrt{a^2-1}}{2}};left(a1right)
Bsqrt{dfrac{a+sqrt{a^2-1}}{2}}-sqrt{dfrac{a-sqrt{a^2-b}}{2}};left(agesqrt{b};bge0right)
Cleft(1+dfrac{a+sqrt{a}}{a+1}right)left(1-dfrac{a-sqrt{a}}{sqrt{a}+1}right);left(age0,ane1right)
Ddfrac{x^2+sqrt{x}}{x-sqrt{x}+1}-dfrac{2x+sqrt{x}}{sqrt{x}};left(x0right)
Đọc tiếp
1. Tìm x để bt có nghĩa
A=\(\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
B=\(\sqrt{\dfrac{2x+3}{x-3}}\)
C=\(\sqrt{-\dfrac{5}{x+2}}\)
D=\(\sqrt{-x}+\dfrac{1}{x+3}\)
2. Rút gọn bt
A=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-1}}{2}};\left(a>1\right)\)
B=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}};\left(a\ge\sqrt{b};b\ge0\right)\)
C=\(\left(1+\dfrac{a+\sqrt{a}}{a+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right);\left(a\ge0,a\ne1\right)\)
D=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}};\left(x>0\right)\)
Tính:
a) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
b) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
Mọi người giúp em với!!!!!!!!!!!!
Tính:
a) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
Bài tập: Thực hiện phép tính
\(\left(\dfrac{1}{3}\sqrt{\dfrac{1}{2}}-\dfrac{2}{3}\sqrt{\dfrac{1}{3}}+\dfrac{2}{7}\sqrt{\dfrac{4}{5}}\right):\left(\dfrac{2}{7}\sqrt{\dfrac{1}{8}}\right)\)
\(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)