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Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{5}=\dfrac{b}{2}=\dfrac{2a-3b}{2\cdot5-3\cdot2}=\dfrac{12}{4}=3\)
Do đó: a=15; b=6
d) Áp dụng t/c dtsbn:
\(\dfrac{a}{5}=\dfrac{b}{2}=\dfrac{2a}{10}=\dfrac{3b}{6}=\dfrac{2a-3b}{10-6}=\dfrac{12}{4}=3\)
\(\Rightarrow\left\{{}\begin{matrix}a=3.5=15\\b=3.2=6\end{matrix}\right.\)
f) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=-\dfrac{z}{2}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{-z}{2}=\dfrac{x+y-z}{5+3+2}=\dfrac{2}{10}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}.5=1\\y=\dfrac{1}{5}.3=\dfrac{3}{5}\\z=\dfrac{1}{5}.\left(-2\right)=-\dfrac{2}{5}\end{matrix}\right.\)
g) \(\dfrac{x}{4}=\dfrac{y}{5}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)
\(\Rightarrow xy=20k^2=500\Rightarrow k=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\\\left\{{}\begin{matrix}x=-20\\y=-25\end{matrix}\right.\end{matrix}\right.\)
4:
a: \(\dfrac{31}{23}-\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\)
\(=\dfrac{31}{23}-\dfrac{7}{32}-\dfrac{8}{23}\)
\(=1-\dfrac{7}{32}=\dfrac{25}{32}\)
b: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+1-1=\dfrac{1}{3}\)
c: \(\left(-30,75\right)+\left(\dfrac{31}{10}-69,25\right)-\left(-6,9\right)\)
\(=-30,75+3,1-69,25+6,9\)
=10-100
=-90
d: \(\left(-34,5\right)\cdot\dfrac{11}{25}-65,5\cdot\dfrac{11}{25}\)
\(=\dfrac{11}{25}\left(-34,5-65,5\right)\)
\(=\dfrac{11}{25}\cdot\left(-100\right)=-44\)
Bài 1:
a: \(\dfrac{1}{6}-0,4\cdot\dfrac{5}{8}+\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{2}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{2-3+6}{12}=\dfrac{5}{12}\)
b: \(\left(-\dfrac{2}{3}\right)^2+\dfrac{1}{6}-\left(-0,5\right)^3\)
\(=\dfrac{4}{9}+\dfrac{1}{6}+\dfrac{1}{8}\)
\(=\dfrac{32+12+9}{72}=\dfrac{53}{72}\)
a: Xét ΔABE và ΔADC có
AB=AD
\(\widehat{BAE}=\widehat{DAC}\)
AE=AC
Do đó: ΔABE=ΔADC
Bài 1 :
Thay x = 2 ; y = -1/2 ta được
\(B=-8+2.4\left(-\dfrac{1}{2}\right)-4.2.\left(\dfrac{1}{4}\right)+2\left(-\dfrac{1}{2}\right)-3\)
\(=-8-4-2-1-3=-18\)
a: \(P=-\left|5-x\right|+2019\le2019\forall x\)
Dấu '=' xảy ra khi x=5
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