\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
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Bài 1: Căn bậc hai
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Tính:
E=(\(\sqrt{18}-3\sqrt{6}+\sqrt{2}\)) \(\sqrt{2}+6\sqrt{3}\)
G=\(\left(2\sqrt{2}-\sqrt{5}+\sqrt{18}\right)\).\(\left(\sqrt{50}+\sqrt{5}\right)\)
H=\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\).\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
a. \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
b. \(3\left(2+\sqrt{x}\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
c. \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
d. \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
thực hiện phép tính
\(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)
\(\left(2+\sqrt{3}-\sqrt{2}\right)\times\left(2-\sqrt{3}-\sqrt{2}\right)\times\left(3+\sqrt{2}\right)\times\sqrt{3-2\sqrt{2}}\)
Dleft ( frac{1}{3sqrt{x}-6} +frac{1}{x-2sqrt{x}}right )left ( frac{1}{6} +frac{1}{2sqrt{x}}right ) Dleft ( frac{1}{3left ( sqrt{x}-2 right )} +frac{1}{sqrt{x}left ( sqrt{x}-2 right )}right ).frac{sqrt{x}+3}{6sqrt{x}} Dfrac{sqrt{x}+3}{3sqrt{x}left ( sqrt{x}-2 right )}.frac{sqrt{x}+3}{6sqrt{x}} Dfrac{left ( sqrt{x}+3 right )^{2}}{18xleft ( sqrt{x}-2 right )} Dfrac{x+6sqrt{x}+9}{18xsqrt{x}-36x}
A/ Đúng
B/ Sai
Đọc tiếp
\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
thực hiện phép tính
A=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
B=\(\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{2}}}\cdot\left(3+\sqrt{5}\right)\)
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right)\times\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right)\times\left(-\sqrt{6}\right)\)
rÚT GỌN: Gsqrt{3+sqrt{5}}+sqrt{7-3sqrt{6}}-sqrt{2}
chững minh : a) 2sqrt{2}left(sqrt{3}-2right)+left(1+2sqrt{2}right)^2-2sqrt[]{6}9
b)left(3+sqrt{5}right)left(sqrt{10}-sqrt{2}right)sqrt{3-sqrt{5}}8
c)sqrt{dfrac{4}{left(2-sqrt{5}right)^2}}-sqrt{dfrac{4}{left(2+sqrt{5}right)^2}}8
giúp mk với tối mai mk nạp rồi
Đọc tiếp
rÚT GỌN: G=\(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{6}}-\sqrt{2}\)
chững minh : a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt[]{6}=9\)
b)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
c)\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
giúp mk với tối mai mk nạp rồi
(1)frac{sqrt{6+4sqrt{2}}}{sqrt{2}} (2)frac{sqrt{3-sqrt{5}}}{sqrt{0.5}} (3)left(sqrt{2}-1right)^2 (4)left(3-2sqrt{2}right).left(3+2sqrt{2}right) (5)sqrt{left(2-sqrt{3}right)}^2-sqrt{left(1-sqrt{3}right)}^2 (6)sqrt{left(sqrt{2}-sqrt{3}right)^2}-sqrt{left(sqrt{2}+sqrt{3}right)}^2 (7)frac{1}{sqrt{3}-1}-frac{1}{sqrt{3}+1} (8)sqrt{3-2sqrt{2}} (9)sqrt{4+2sqrt{3}}-sqrt{4-2sqrt{3}} (10)sqrt{2020+2sqrt{2019}}-sqrt{2020-2sqrt{2019}} (11)sqrt{7+2sqrt{12}} Các bạn giúp mình với ,Mình xin cảm ơn trước
Đọc tiếp
(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (10)\(\sqrt{2020+2\sqrt{2019}}-\sqrt{2020-2\sqrt{2019}}\) (11)\(\sqrt{7+2\sqrt{12}}\) Các bạn giúp mình với ,Mình xin cảm ơn trước