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\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(A=1-1+1=1\)
\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)
\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)
\(A=-1+1=0\)
\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=-36-\dfrac{1}{4}\)
B = \(-\dfrac{145}{4}\)
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a: \(A=\dfrac{19}{5}x^4y^3\)
b: Hệ số là 19/5
Bậc là 7
c: \(A=\dfrac{19}{5}\cdot1^4\cdot2^3=\dfrac{152}{5}\)
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a,
\(A=\frac{19}{5}xy^2\left(x^3y\right)\left(-3x^{13}y^5\right)^0\)
\(A=\frac{19}{5}xy^2\left(x^3y\right).1\)
\(A=\frac{19}{5}\left(x.x^3\right)\left(y^2.y\right)\)
\(A=\frac{19}{5}x^4y^3\)
b, Hệ số: \(\frac{19}{5}\) ; Bậc: 7
c, Giá trị đơn thức \(A=\frac{19}{5}x^4y^3\) tại x=1, y=2
\(A=\frac{19}{5}.1^4.2^3\)
\(A=\frac{152}{5}=3\frac{2}{5}\)
Vậy...
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a, \(\frac{3}{5}+\frac{-1}{3}\)
=\(\frac{9}{15}+\frac{-5}{15}\)
=\(\frac{4}{15}\)
b,\(\frac{-2}{13}+\frac{-11}{26}\)
=\(\frac{-4}{26}+\frac{-11}{26}\)
=\(\frac{-15}{26}\)
c, \(-2+\frac{-5}{8}\)
=\(\frac{-16}{8}+\frac{-5}{8}\)
=\(\frac{-21}{8}\)
d,\(\frac{13}{30}-\frac{1}{5}\)
=\(\frac{13}{30}-\frac{6}{30}\)
=\(\frac{7}{30}\)
e,\(\frac{2}{21}-\frac{-1}{28}\)
=\(\frac{56}{588}-\frac{-21}{588}\)
=\(\frac{77}{588}=\frac{11}{84}\)
bài dễ mà <3