Tại sao 2 - 3 - 4 + 5 = 0 . ( Tính từng bước một ) .
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\(\dfrac{1}{\sqrt{3-2\sqrt{2}}}+\dfrac{1}{\sqrt{5-2\sqrt{6}}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}}+\dfrac{1}{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{2}-1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)
\(=\dfrac{1}{\left|\sqrt{2}-1\right|}+\dfrac{1}{\left|\sqrt{3}-\sqrt{2}\right|}\)
\(=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}\right)^2-1}+\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}\)
\(=\sqrt{2}+1+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}+1\)
\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(=\frac{1}{2}(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12})\)
\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{10}-\frac{1}{12})\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{12}\right)=\frac{1}{2}\cdot\frac{5}{12}=\frac{5}{24}\)
c) \(\dfrac{11}{10}-\dfrac{-7}{24}=\dfrac{11}{10}+\dfrac{7}{24}=\dfrac{167}{120}\)
e) \(\dfrac{-8}{3}\cdot\dfrac{15}{7}=\dfrac{-120}{21}=\dfrac{-40}{7}\)
f) \(\dfrac{-2}{5}\cdot4\dfrac{1}{2}=\dfrac{-2}{5}\cdot\dfrac{9}{2}=-\dfrac{9}{5}\)
g) \(\dfrac{5}{3}:\dfrac{5}{-3}=\dfrac{5}{3}:\dfrac{-5}{3}=\dfrac{5}{3}\cdot\dfrac{-3}{5}=-1\)
h) \(\dfrac{5}{4}:\left(-9\right)=\dfrac{5}{4}:\dfrac{-9}{1}=\dfrac{5}{4}\cdot\dfrac{-1}{9}=-\dfrac{5}{36}\)
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)
Ta có: 2-3-4+5
= (-1)-4+5
= (-5)+5
= 0
Vậy 2-3-4+5=0
2 - 3 - 4 + 5 = 0 . Vì :
2 - 3 - 4 + 5
= ( - 1 ) - 4 + 5
= ( - 5 ) + 5
= 0
Nhớ tick cho mik nha !!!