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\(\dfrac{3}{8}+\dfrac{6}{15}:\dfrac{3}{5}\\ =\dfrac{3}{8}+\dfrac{2}{5}.\dfrac{5}{3}\\ =\dfrac{3}{8}+\dfrac{2}{3}\\ =\dfrac{9}{24}+\dfrac{16}{24}\\ =\dfrac{25}{24}\)
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a, \(y-\frac{1}{6}=\frac{5}{6}\)
\(y=\frac{5}{6}+\frac{1}{6}\)
\(y=1\)
b, 1295.23-16839
=29785-16839
=12946
_HT_
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\(x^2+4y^2+z^2-2x+8y-6x+15=0\)
<=> \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1=0\)
mà \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\)≥0
=> \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)≥1
=> ko có giá trị nào của x,y,z thỏa mãn
\(A=\dfrac{1}{x^2-4x+9}=\dfrac{1}{\left(x-2\right)^2+5}\)
mà (x+2)2≥0
=> (x+2)2+5≥5
=> \(\dfrac{1}{\left(x-2\right)^2+5}\)≤ 1/5
=> Max A = 1/5 dấu ''='' xảy ra khi x=2
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Thay \(a=\dfrac{16}{15}\) vào biểu thức ta được
\(\dfrac{3}{4}\) x \(\dfrac{16}{15}+\dfrac{6}{5}=\dfrac{4}{5}+\dfrac{6}{5}=\dfrac{4+6}{5}=\dfrac{10}{5}=2\)
Vậy với \(a=\dfrac{16}{15}\) giá trị biểu thức là 2
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a, \(\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{1}{6}=\dfrac{1}{48}\)
b, \(\dfrac{1}{2}.\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{1}{8}.6=\dfrac{6}{8}=\dfrac{3}{4}\)
\(a,\dfrac{1}{2}\times\dfrac{1}{4}\times\dfrac{1}{6}=\dfrac{1}{8}\times\dfrac{1}{6}=\dfrac{1}{48}\)
\(b,\dfrac{1}{2}\times\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{6}{8}=\dfrac{3}{4}\)