Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)
\(=\frac{24}{2}\)
\(=12\)
\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}+2-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=1+\sqrt{2}\)
\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{2\left(2-\sqrt{3}\right)}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(A=3-1=2\)
P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé
a: \(=3\cdot7\sqrt{3}+2\cdot6\sqrt{3}-4\cdot4\sqrt{3}-11\sqrt{3}\)
\(=21\sqrt{3}+12\sqrt{3}-16\sqrt{3}-11\sqrt{3}\)
\(=6\sqrt{3}\)
b: \(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=3-\sqrt{5}+\sqrt{5}-1\)
=2
c: \(=\left(4-\sqrt{3}\right)\sqrt{\left(4+\sqrt{3}\right)^2}\)
\(=\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)\)
=16-3
=13
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
Giải:
a) \(5\sqrt{\left(-2\right)^4}\)
\(=5\sqrt{2^4}\)
\(=5\sqrt{\left(2^2\right)^2}\)
\(=5.2^2\)
\(=20\)
b) \(-4\sqrt{\left(-3\right)^6}\)
\(=-4\sqrt{3^6}\)
\(=-4\sqrt{\left(3^3\right)^2}\)
\(=-4.3^3\)
\(=-108\)
c) \(\sqrt{\sqrt{\left(-5\right)^8}}\)
\(=\sqrt{\sqrt{5^8}}\)
\(=\sqrt{\sqrt{\left(5^4\right)^2}}\)
\(=\sqrt{5^4}\)
\(=5^2=25\)
d) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\)
\(=2\sqrt{5^6}+3\sqrt{2^8}\)
\(=2.5^3+3.2^4\)
\(=250+48\)
\(=298\)
Vậy ...
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)