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Bạn ơi bạn có thể giải cách lớp 7 ko ạ tại đường trung bình em chx hc ý ạ
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d)\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4=\dfrac{1}{3^{50}}.\left(-3^{50}\right)-\dfrac{1}{6}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
e) \(=\left(4.\dfrac{3}{4}-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\left(3-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\dfrac{5}{2}.\dfrac{6}{5}-17=3-17=-14\)
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\(=\dfrac{3}{4}.\dfrac{-8}{5}+\dfrac{1}{4}=-\dfrac{6}{5}+\dfrac{1}{4}=-\dfrac{19}{20}\)
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\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right):\dfrac{2}{9}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right):\dfrac{2}{3}=\)
\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)\cdot\dfrac{3}{2}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\cdot\dfrac{3}{2}=\)
\(\left[\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\right]\cdot\dfrac{3}{2}=\)
\(\left[\dfrac{1}{2020}-\dfrac{2018}{2019}-\dfrac{2019}{2020}-\dfrac{1}{2019}\right]\cdot\dfrac{3}{2}=\)
\(0\)
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{xy+1}{10}=\dfrac{yz+2}{27}=\dfrac{xz+2}{19}=\dfrac{xy+1+yz+2+xz+2}{10+27+19}=\dfrac{55}{56}\)
=>xy+1=550/56; yz+2=1485/56; xz+2=1045/56
=>xy=247/28; yz=1373/56; xz=933/56
=>xyz=60,03 hoặc xyz=-60,03
TH1: xyz=60,03
=>z=6,81; x=2,45; y=3,60
TH2: xyz=-60,03
=>z=-6,81; x=2,45; y=3,60