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Simplifiez l'expression \(\displaystyle{\sqrt[n]{\frac{a^{2n+1}}{b^{n+1}}}}\).
\(\displaystyle{\frac{b^{n^2+n}}{a^{2n^2+n}}}\)
\(\displaystyle{\frac{a^2\sqrt[n]{a}}{b\sqrt[n]{b}}}\)
\(\displaystyle{\frac{b^{2n+1}}{a^{3n+1}}}\)
\(\displaystyle{\frac{a^2+\sqrt[n]{a}}{b+\sqrt[n]{b}}}\)
Calculez \(\displaystyle{\frac{\frac{6}{x}+3}{\frac{5}{x}-\frac{2}{x}}}\).
\(3\)
\(2+x\)
\(\dfrac{3x}{x+1}\)
\(0\)
Calculez \(\dfrac{3}{4}-\left(\dfrac{1}{4}-\left(\dfrac{2}{3}-\dfrac{3}{4}\right)\right)\)
\(\dfrac{1}{12}\)
\(\dfrac{11}{12}\)
\(\dfrac{5}{12}\)
\(\dfrac{7}{12}\)
Calculez \(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)/\left(\dfrac{1}{x^2}-\dfrac{1}{y^2}\right)\).
\(\dfrac{1}{y-x}\)
\(\dfrac{x^2y+x^3-y^3-xy^2}{xy}\)
\(x-y\)
\(\dfrac{xy}{y-x}\)
Calculez \(\left(\dfrac{x-y}{x+y}\right)/\left(\dfrac{x^2-y^2}{x^2+2xy+y^2}\right)\).
\(\dfrac{x+y}{x-y}\)
\(\left( \dfrac{x-y}{x+y}\right)^2\)
1
\(-1\)
Calculez \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\).
\(a-b\)
\(a+b\)
\(\sqrt{a}+\sqrt{b}\)
\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{a}}\)
Calculez \(\dfrac{\sqrt{14}+\sqrt{15}}{\sqrt{7}-\sqrt{5}}\).
\(\dfrac{29}{2}\)
\(\dfrac{7\sqrt{2}+\sqrt{70}+\sqrt{105}+5\sqrt{3}}{12}\)
\(\dfrac{7\sqrt{2}+\sqrt{70}+\sqrt{105}+5\sqrt{3}}{2}\)
\(\dfrac{\sqrt{21}+\sqrt{19}+\sqrt{22}+\sqrt{20}}{2}\)
Le double du carré de l'opposé de 0,4 est
-0,32
-0,64
0,64
0,32
Calculez \(\dfrac{\sqrt{2}}{2-2\sqrt{2}}\).
\(-\dfrac{\sqrt{2}}{2}-1\)
\(\dfrac{\sqrt{2}}{6}+\dfrac{1}{3}\)
\(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)
\(-\sqrt{2}\)
Le carré de l'opposé du double de 0,4 est
\(-0,64\)
\(0,64\)
\(0,16\)
\(-0,32\)
