Trục căn thức ở mẫu với giả thiết $a, b, x, y$ là những số dương:
a) $\dfrac{a}{\sqrt{a}}$;
b) $\dfrac{a}{\sqrt{a b}}$;
c) $\dfrac{x}{\sqrt{3 x^{3}}}$;
d) $\dfrac{4 y^{2}}{\sqrt{2 y^{5}}}$.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
) x y \sqrt{\dfrac{x}{y}}=x y \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x y}{y} \sqrt{x y}=x \sqrt{x y} .xyyx=xyy2xy=yxyxy=xxy.
b) \dfrac{x}{y} \sqrt{\dfrac{x}{y}}=\dfrac{x}{y} \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x}{y^{2}} \sqrt{x y} .yxyx=yxy2xy=y2xxy.
c) \sqrt{\dfrac{1}{a}+\dfrac{1}{a^{2}}}=\sqrt{\dfrac{a+1}{a^{2}}}=\dfrac{\sqrt{a+1}}{a} .a1+a21=a2a+1=aa+1.
d) \sqrt{\dfrac{4 x^{3}}{25 y}}=\sqrt{\dfrac{4 x^{2} x y}{25 y^{2}}}=\dfrac{2 x}{5 y} \sqrt{x y} .25y4x3=25y24x2xy=5y2xxy.
e) 2 a b \sqrt{\dfrac{3}{a b}}=2\sqrt{\dfrac{3(ab)^2}{ab}}=2\sqrt{3ab}2abab3=2ab3(ab)2=23ab.
a) \(\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}.\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\)
b) \(\sqrt{\frac{3a}{5b}}=\frac{\sqrt{3a}}{\sqrt{5b}}=\frac{\sqrt{3a}.\sqrt{5b}}{5b}=\frac{\sqrt{15ab}}{5b}\left(a;b>0\right)\)
c) \(\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{12}}=\frac{\sqrt{5}.\sqrt{12}}{12}=\frac{\sqrt{60}}{12}=\frac{2\sqrt{15}}{12}=\frac{\sqrt{15}}{6}\)
d) \(\sqrt{\frac{5x}{18y}}=\frac{\sqrt{5x}}{\sqrt{18y}}=\frac{\sqrt{5x}}{\sqrt{3^2.2y}}=\frac{\sqrt{5x}}{3\sqrt{2y}}\)
\(=\frac{\sqrt{5x}.\sqrt{3y}}{3.2y}=\frac{\sqrt{15xy}}{6xy}\)
tham khảo
KỲ QUAN VỊNH HẠ LONG, QUẢNG NINH.
a) Ta sắp xếp theo thứ tự tăng dần như sau:
\(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)
b) Ta sắp xếp theo thứ tự tăng dần như sau:
\(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)
a, \(\frac{a}{\sqrt{a}}=\sqrt{a}\)
b, \(\frac{a}{\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
c, \(\frac{x}{\sqrt{3x^3}}=\frac{x}{x\sqrt{3x}}=\frac{1}{\sqrt{3x}}=\frac{\sqrt{3x}}{3x}\)
d, \(\frac{4y^2}{\sqrt{2y^5}}=\frac{4y^2}{y^2\sqrt{2y}}=\frac{4}{\sqrt{2y}}=\frac{4\sqrt{2y}}{2y}=\frac{2\sqrt{2y}}{y}\)
a)\(\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}\) b) \(\dfrac{a}{\sqrt{ab}}=\dfrac{a\sqrt{ab}}{\left(\sqrt{ab}\right)^2}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b}\) c) \(\dfrac{x}{\sqrt{3x^3}}=\dfrac{x\sqrt{3x}}{\sqrt{3x^3.\sqrt{3x}}}=\dfrac{x\sqrt{3x}}{\left(\sqrt{3x^2}\right)^2}=\dfrac{x\sqrt{3x}}{\left(3x^2\right)^2}=\dfrac{x\sqrt{3x}}{3x^2}=\dfrac{\sqrt{3x}}{3x}\)