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a; \(\dfrac{1}{2}-\dfrac{-3}{6}+\dfrac{5}{3}-\dfrac{9}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{5}{3}-\dfrac{3}{4}\)
\(=1-\dfrac{3}{4}+\dfrac{5}{3}=\dfrac{1}{4}+\dfrac{5}{3}=\dfrac{3+20}{12}=\dfrac{23}{12}\)
b: \(=\dfrac{3}{11}\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}\cdot\dfrac{-6-16}{9}=\dfrac{3}{11}\cdot\dfrac{-22}{9}=\dfrac{-2}{3}\)
c: \(=1-3+\dfrac{1}{4}=-2+\dfrac{1}{4}=-\dfrac{7}{4}\)
1.
\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{1}{6}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{1}{9}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{9}\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{10}{27}\right\}\)
=\(\dfrac{2}{3}.\dfrac{8}{27}\)
=...
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)
\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)
\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)
\(\Rightarrow\left(a+b+c\right)^2=60\)
\(\Rightarrow a+b+c=\pm\sqrt{60}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)
Các câu sau làm tương tự
b. \(ab=\dfrac{3}{5};bc=\dfrac{4}{5};ac=\dfrac{3}{4}\)
\(\Rightarrow ab\cdot bc\cdot ac=\dfrac{9}{25}\Rightarrow\left(abc\right)^2=\dfrac{9}{25}\Rightarrow abc=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\dfrac{3}{5}:bc=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\\b=\dfrac{3}{5}:ac=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\c=\dfrac{3}{5}:ab=\dfrac{3}{5}:\dfrac{3}{5}=1\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\dfrac{3}{5}:\dfrac{4}{5}=-\dfrac{3}{4}\\b=-\dfrac{3}{5}:\dfrac{3}{4}=-\dfrac{4}{5}\\c=-\dfrac{3}{5}:\dfrac{3}{5}=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy......................
Ta có: \(a+b+c=1 \)
\(\Leftrightarrow(a+b+c)^2=1 \)
\(\Leftrightarrow ab+bc+ca=0 (1) \)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{(x+y+z)}{\left(a+b+c\right)}=x+y+z\)
\(\Leftrightarrow x=a\left(x+y+z\right)\)
\(\Leftrightarrow y=b.\left(x+y+z\right)\)
\(\Leftrightarrow z=c.\left(x+y+z\right)\)
\(\Rightarrow xy+yz+zx=ab.\left(x+y+z\right)^2+bc.\left(x+y+z\right)^2+ca.\left(x+y+z\right)^2\)
\(\Leftrightarrow xy+yz+zx=\left(ab+bc+ca\right).\left(x+y+z\right)^2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra: \(xy+yz+zx=0\)
Đề ảo tek.Sửa đề.
\(\left\{{}\begin{matrix}a+b+c=5\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=25\\\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\bc+ac+ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\2bc+2ac+2ab=0\end{matrix}\right.\)
\(\Leftrightarrow a^2+b^2+c^2+2ab-2ab+2bc-2bc+2ac-2ac=25\)
\(\Leftrightarrow a^2+b^2+c^2=25\)
a)
\(\left[\dfrac{-2}{3}+0,5:\left(\dfrac{-3}{2}\right)^2\right]+\left[1\dfrac{1}{5}-1,4.\dfrac{5}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{1}{2}:\dfrac{9}{4}\right]+\left[\dfrac{6}{5}-\dfrac{7}{5}.\dfrac{5}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{1}{2}.\dfrac{4}{9}\right]+\left[\dfrac{6}{5}-\dfrac{7}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{2}{3}\right]+\left[\dfrac{36}{30}-\dfrac{35}{30}+6\right]\\ =0+\left[\dfrac{1}{30}+6\right]\\ =6\dfrac{1}{30}\)
b)
\(\left(-0,2\right)^2.5-\dfrac{8^2.9^5}{3^9.4^3}\\ =0,4.5-\dfrac{\left(2^3\right)^2.\left(3^2\right)^5}{3^9.\left(2^2\right)^3}\\ =2-\dfrac{2^6.3^{10}}{3^9.2^6}\\ =2-\dfrac{1.3}{1.1}\\ =2-3\\ =-1\)
mọi người lưu ý nhé chỗ cuối phép tính là nhoặc to nha