Liên hệ giữa phép chia và phép khai phương
Câu 1 : Tính:\(\begin{array}{l}a)\,\,\sqrt {\frac{{289}}{{225}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\sqrt {2\frac{{14}}{{25}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)\,\,\sqrt {\frac{{0,25}}{9}} \\d)\,\,\sqrt {1\frac{{9}}{16}.5\frac{4}{9}.0,01} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e)\,\,\sqrt {\frac{{{{149}^2} - {{76}^2}}}{{{{457}^2} - {{384}^2}}}} \end{array}\)
A \(\begin{array}{l}
a)\,\,\,\frac{{27}}{{5}} & & & b)\,\,\frac{8}{5}\\
c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\
e)\,\,\frac{{15}}{{29}}
\end{array}\)
B \(\begin{array}{l}
a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\
c)\,\,\frac{1}{6} & & & d)\,\,\frac{17}{{4}}\\
e)\,\,\frac{{15}}{{29}}
\end{array}\)
C \(\begin{array}{l}
a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\
c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\
e)\,\,\frac{{25}}{{9}}
\end{array}\)
D \(\begin{array}{l}
a)\,\,\,\frac{{17}}{{15}} & & & b)\,\,\frac{8}{5}\\
c)\,\,\frac{1}{6} & & & d)\,\,\frac{7}{{24}}\\
e)\,\,\frac{{15}}{{29}}
\end{array}\)
Câu 2 :
Tính:\(\begin{array}{l}
a)\,\,\frac{x}{y}\sqrt {\frac{{{x^2}}}{{{y^4}}}} \,\,\left( {x > 0,y \ne 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \,\,\left( {x < 0,y > 0} \right)\\
c)\,\,a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}} \,\,\left( {a < 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}} \,\,\left( {a \ge - \frac{3}{2},\,\,\,b < 0} \right)
\end{array}\)
A \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, \frac{{25{x^2}}}{y}\\
c)\,\, - \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ - b}}
\end{array}\)
B \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, - \frac{{25{x^2}}}{y^2}\\
c)\,\, - \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ - b}}
\end{array}\)
C \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, - \frac{{25{x^2}}}{y}\\
c)\,\, \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ - b}}
\end{array}\)
D \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, - \frac{{25{x^2}}}{y}\\
c)\,\, -\sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ b}}
\end{array}\)
Câu 3 : Rút gọn biểu thức\(\begin{array}{l}a)\,\,\left( {5\sqrt {48} + 4\sqrt {27} - 2\sqrt {12} } \right):\sqrt 3 \\b)\,\,\left( {\sqrt {{a^2} - {b^2}} + \sqrt {\left( {a + b} \right).b} } \right):\sqrt {a + b} \,\,\,\left( {a > b > 0} \right).\end{array}\)
A \(\begin{array}{l}
a)\,\,\sqrt{3}\\
b)\,\,\sqrt {a - b} + \sqrt b
\end{array}\)
B \(\begin{array}{l}
a)\,\,28\\
b)\,\,\sqrt {a - b} + \sqrt b
\end{array}\)
C \(\begin{array}{l}
a)\,\,28\\
b)\,\,\sqrt {a + b} + \sqrt b
\end{array}\)
D \(\begin{array}{l}
a)\,\,3\sqrt{3}\\
b)\,\,\sqrt {a - b} + \sqrt b
\end{array}\)
Câu 4 : Tìm \(x\) biết:\(a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3\)
A \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{- 7 \right\}.
\end{array}\)
B \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 7 \right\}.
\end{array}\)
C \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 2 \right\}.
\end{array}\)
D \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 2;\,7 \right\}.
\end{array}\)
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