Cho a+b-ab=-1 và \(a^2+b^2=13\). Tìm \(P=|a^3-b^3|\)
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\(a^2+b^2=13\Leftrightarrow a^2+b^2+2ab-2ab=13\Leftrightarrow\left(a+b\right)^2-2ab=13\)
Mà \(a+b-ab=-1\Leftrightarrow ab=a+b+1\)Thay vào phương trình trêm ta có:
\(\left(a+b\right)^2-2\left(a+b+1\right)=13\)
<=> \(\left(a+b\right)^2-2\left(a+b\right)+1=16\)
<=> \(\left(a+b+1\right)^2=4^2\)
<=> \(a+b+1=\pm4\)=> \(ab=\pm4\)
Ta lại có: \(a^2+b^2=13\Leftrightarrow\left(a-b\right)^2+2ab=13\)
+) Với ab=4
thay vào ta có: \(\left(a-b\right)^2+8=13\Leftrightarrow\left(a-b\right)^2=5\Leftrightarrow\left|a-b\right|=\sqrt{5}\)
=> \(P=\left|a^3-b^3\right|=\left|\left(a-b\right)\left(a^2+b^2+ab\right)\right|=\left|a-b\right|\left|a^2+b^2+ab\right|\)
\(=\sqrt{5}\left(13+4\right)=17\sqrt{5}\)
+) Với ab=-4 . Em làm tương tự nhé!
a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\)
\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\)
b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)
\(\dfrac{a}{b}\) = \(\dfrac{4}{9}\)
c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{13}{21}\)
d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)
bài này thử là nhanh nhất (hi hi , mình đùa vui thôi chứ minh ko bít làm)
a, \(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)
\(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\) = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)
\(\dfrac{4}{7}\) \(\times\) \(\dfrac{a}{b}\) = \(\dfrac{8}{21}\)
\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\): \(\dfrac{4}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\)
b, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{13}{21}\)