\(=\left(6\sqrt{2}-5\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=2-1=1\)
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Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
Rg
a) \(\dfrac{(3\sqrt{3}+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt{2}}\)
b)\((\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}})(\sqrt{6}+11)\)
a)\(\sqrt{50}-\sqrt{3}.\sqrt{6}+\frac{\sqrt{22}}{\sqrt{11}}\)
b) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\sqrt{7+4\sqrt{3}}\)
Thực hiên các phép tính:
a) 2sqrt{75}-4sqrt{12}-3sqrt{50}-sqrt{72}
b) sqrt{ab}+2sqrt{dfrac{b}{a}}-sqrt{dfrac{b}{a}}+sqrt{dfrac{1}{ab}}left(a0,b0right)
c) 5sqrt{a}-3sqrt{25a^3}+2sqrt{36ab^2}-2sqrt{9a}left(a0,b0right)
d) dfrac{2sqrt{6}-2sqrt{3}}{sqrt{2}-1}-dfrac{3+sqrt{3}}{sqrt{3}}+sqrt{27}
e) dfrac{1}{1+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+...+dfrac{1}{sqrt{99}+sqrt{100}}
f) dfrac{1}{2}sqrt{48}-2sqrt{75}-sqrt{54}+5sqrt{1dfrac{1}{3}}
g) 0,1sqrt{200}+2sqrt{0,08}+0,4sqrt{50}
Đọc tiếp
Thực hiên các phép tính:
a) \(2\sqrt{75}-4\sqrt{12}-3\sqrt{50}-\sqrt{72}\)
b) \(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{1}{ab}}\left(a>0,b>0\right)\)
c) \(5\sqrt{a}-3\sqrt{25a^3}+2\sqrt{36ab^2}-2\sqrt{9a}\left(a>0,b>0\right)\)
d) \(\dfrac{2\sqrt{6}-2\sqrt{3}}{\sqrt{2}-1}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\sqrt{27}\)
e) \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
f) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\dfrac{1}{3}}\)
g) \(0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
tìm x :
a, \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25+50}=6\)
b,\(\sqrt{4x^2-4x+1}=x+2\)
c, \(3\sqrt[3]{x-3}+4\sqrt[3]{8x-24}-\dfrac{1}{3}\sqrt[3]{2x+81}=20\)
d, \(\dfrac{x}{2}-\sqrt{x-1}=2\)
\(\left(\frac{1}{2}.\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{4.5}+\frac{2}{5}\sqrt{50}\right):\frac{4}{5}\sqrt{\frac{1}{8}}\)
Tính
\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
Bài 1.Thực hiện phép tính
1) 2 / 3+2√2 + 2 / 3-2√2
2) A= √ 15-6√6 + √ 15- 6√6
B2: Rút gọn bt
P= ( x√x+1 / √x+1 -√x) : (x-1 / √x+1)
*Dấu"/" là dấu phần ạ*
Mong mn giải giúp mk bài trên.Thx
1.3+√3/√5 - 2/√3-1 2.√12-√5/√2-1 -1/√5-2 3.2√27-6√1/3+1/2+√3-9/√3
Thực hiện phép tính sau:
K=\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)