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\(a,\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)
\(=\dfrac{9}{3\left(\sqrt{7}-4\right)}+\dfrac{\left(\sqrt{7}-4\right)\left(\sqrt{7}+4\right)}{3\left(\sqrt{7}-4\right)}\)
\(=\dfrac{9+7-16}{3\left(\sqrt{7}-4\right)}\)
\(=0\)
\(b,\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\right]\cdot2\sqrt{3}\)
\(=\left(\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\right)\cdot2\sqrt{3}\)
\(=\left(\sqrt{2}+\sqrt{3}-\sqrt{2}\right)\cdot2\sqrt{3}\)
\(=\sqrt{3}\cdot2\sqrt{3}\)
\(=6\)
#\(Toru\)
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
a: \(=\dfrac{2\sqrt{2}+3+2\sqrt{2}-3}{8-9}\)
\(=\dfrac{4\sqrt{2}}{-1}=-4\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}-\sqrt{7}\right)+\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)}{8-7}\)
\(=4-\sqrt{14}+4+\sqrt{14}=8\)
c: \(=\dfrac{2+\sqrt{5}-2\left(2-\sqrt{5}\right)}{-1}=\dfrac{2+\sqrt{5}-4+2\sqrt{5}}{-1}\)
\(=-3\sqrt{5}+2\)
a) \(=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(=\dfrac{3\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=3\sqrt{2}+3+\sqrt{2}=3+4\sqrt{2}\)
c) \(=\dfrac{3\left(\sqrt{5}+2\right)-3\left(\sqrt{5}-2\right)}{5-4}=3\sqrt{5}+6-3\sqrt{5}+6=12\)
a)\(\dfrac{\sqrt{21}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}+\dfrac{4\left(5+\sqrt{21}\right)}{4}-\dfrac{\sqrt{3}.\sqrt{2}.\sqrt{7}}{\sqrt{3}}\)=\(5+2\sqrt{21}-\sqrt{14}\)
c) (\(\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}\))+\(\sqrt{2}\left(\sqrt{2-\sqrt{3}}\right)\)=1+\(\sqrt{2\sqrt{2}-\sqrt{6}}\)
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)
\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)
\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn
a: \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}\cdot\sqrt{a}-\sqrt{a}}{-\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{-\left(\sqrt{a}-1\right)}=-\sqrt{a}\)
b: \(\dfrac{2+\sqrt{3}}{2-\sqrt{7}}=\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\)
\(=\dfrac{-\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{3}\)
\(=\dfrac{-4-2\sqrt{7}-2\sqrt{3}-\sqrt{21}}{3}\)
c: \(3xy\cdot\sqrt{\dfrac{2}{xy}}=\dfrac{3xy}{\sqrt{xy}}\cdot\sqrt{2}=3\sqrt{2}\cdot\sqrt{xy}\)
d:
\(\dfrac{3}{\sqrt[3]{3}+\sqrt[3]{2}}=\dfrac{3\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{3+2}\)
\(=\dfrac{3}{5}\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)\)
e:
\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
\(=\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{5}{2-\sqrt{3}}-\dfrac{6}{3-\sqrt{3}}\)
\(=2\left(\sqrt{3}+1\right)-\dfrac{5\left(2+\sqrt{3}\right)}{4-3}-\dfrac{6\left(3+\sqrt{3}\right)}{6}\)
\(=2\sqrt{3}+2-10-5\sqrt{3}-3-\sqrt{3}\)
\(=-4\sqrt{3}-11\)
f:
\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\)
\(=\dfrac{\sqrt{5}-1}{5-1}+\dfrac{\sqrt{9}-\sqrt{5}}{9-5}+\dfrac{\sqrt{13}-\sqrt{9}}{13-9}\)
\(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-\sqrt{9}+\sqrt{13}}{4}=\dfrac{\sqrt{13}-1}{4}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\\ =\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\\ =-\sqrt{a}\\ \dfrac{2+\sqrt{3}}{2-\sqrt{7}}\\ =\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\\ =\dfrac{4+2\sqrt{7}+2\sqrt{3}+\sqrt{21}}{-3}\\\)
\(3xy\sqrt{\dfrac{2}{xy}}\\ =\sqrt{\dfrac{\left(3xy\right)^2\cdot2}{xy}}\\ =\sqrt{\dfrac{9x^2y^2\cdot2}{xy}}\\ =\sqrt{9xy\cdot2}\\ =\sqrt{18xy}\)
\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{5\left(\sqrt{3}+2\right)}{3-4}+\dfrac{6\left(\sqrt{3}+3\right)}{3-9}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{2}-\dfrac{5\left(\sqrt{3}+2\right)}{-1}+\dfrac{6\left(\sqrt{3}+3\right)}{-6}\\ =2\sqrt{3}+2+5\sqrt{3}+10-\sqrt{3}-3\\ =6\sqrt{3}+9\)
\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\\ =\dfrac{1-\sqrt{5}}{1-5}+\dfrac{\sqrt{5}-\sqrt{9}}{5-9}+\dfrac{\sqrt{9}-\sqrt{13}}{9-13}\\ =\dfrac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}}{-4}\\ =\dfrac{1-\sqrt{13}}{-4}\)
`# gvy`
a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\) =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\) - \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+ \(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}.21a}\) - \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\) - \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+ \(\sqrt{21a}\)
=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\) + \(\sqrt{21a}\)
=\(\frac{-10}{21}\sqrt{21a}\)
b)
N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)
=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)
=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)
=\(\frac{1}{6}\sqrt{6x}\)
em lớp 8 nene làm theo cách hiểu thôi ạ
Ta có \(\left\{{}\begin{matrix}a+b=\dfrac{-2+\sqrt{3}}{3}+\dfrac{-2-\sqrt{3}}{3}=-\dfrac{4}{3}\\ab=\dfrac{\left(-2+\sqrt{3}\right)\left(-2-\sqrt{3}\right)}{9}=\dfrac{1}{9}\end{matrix}\right.\)
\(\left(a+b\right)^2=a^2+b^2+2ab=16\\ \Leftrightarrow a^2+b^2=\dfrac{16}{9}-2\cdot\dfrac{1}{9}=\dfrac{14}{9}\left(1\right)\\ \left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3+\dfrac{1}{3}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3=-\dfrac{64}{27}+\dfrac{4}{9}=-\dfrac{52}{27}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(a^2+b^2\right)\left(a^3+b^3\right)=a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{14}{9}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5+\dfrac{1}{81}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5=-\dfrac{728}{243}+\dfrac{4}{243}=-\dfrac{724}{243}\left(3\right)\)
\(\left(1\right)\left(3\right)\Rightarrow\left(a^2+b^2\right)\left(a^5+b^5\right)=a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{14}{9}\cdot\left(-\dfrac{724}{243}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7+\dfrac{1}{81}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7=-\dfrac{10136}{2187}-\dfrac{52}{2187}=-\dfrac{10188}{2187}=\dfrac{1132}{243}\)