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a/\(-5,94-13,78=-19,72\)
b/\(\dfrac{3}{4}-\dfrac{5}{4}:\dfrac{-10}{16}\)
\(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{5}\)
\(=\dfrac{3}{4}+\dfrac{8}{4}\)
\(=\dfrac{11}{4}\)
c/\(\dfrac{12}{17}\cdot\dfrac{5}{9}-\dfrac{5}{9}\cdot\dfrac{29}{17}+1\dfrac{5}{9}\)
\(=\left(\dfrac{12}{17}-\dfrac{29}{17}\right)\dfrac{5}{9}+1\dfrac{5}{9}\)
\(=-1\cdot\dfrac{5}{9}+1\dfrac{5}{9}\)
\(=\dfrac{-5}{9}+1\dfrac{5}{9}\)
\(=1\)
#DatNe |
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4.2\(^x\)+3.2\(^{x+1}\)=80
4.2\(^x\)+3.2\(^x\).2\(^1\)=80
2\(^x\).(4+3.2)=80
2\(^x\).10=80
2\(^x\)=8
⇒x=3 vì 2\(^3\)=8
= 22 .2X+3.2X.21=80
=4.2X+3.2.2X
= 4.2X+6.2x
=2X.(4+6)
= 2X.10=80
2X=80:10
2X= 8
X= 8:2
X=4
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\(d,3.2^{x+2}-9.2^x=96\\ 3.2^2.2^x-9.2^x=96\\ 12.2^x-2^x.9\\ 2^x.\left(12-9\right)=96\\ 2^x.3=96\\ 2^x=96:3\\ 2^x=32\\ 2^x=2^5\\ x=5\)
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\(\Leftrightarrow2^x\cdot8-3\cdot2^x\cdot2=32\)
=>x+1=5
hay x=4
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\(3\cdot2x=15\cdot2^{12}+9\cdot8^4\)
=>\(3\cdot2x=15\cdot2^{12}+9\cdot2^{12}\)
=>\(3\cdot2x=24\cdot2^{12}\)
=>\(2x=8\cdot2^{12}=2^{15}\)
=>\(x=2^{14}\)
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Lời giải:
$2^x+4.2^{x+1}+3.2^{x+2}=92$
$2^x+8.2^x+12.2^x=92$
$2^x(1+8+12)=92$
$2^x.21=92$
$2^x=\frac{92}{21}$ không là số tự nhiên. Đề có vẻ sai.
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a) \(3^x:3^2=243\)
\(3^{x-2}=3^5\)
=> x - 2 = 5
=> x = 7
b) \(6.2^x+3.2^x=2^x.\left(6+3\right)=2^x.9=9.29=261\)
=> 2x = 9
=> x không tồn tại vì 2 khi nâng lên lũy thừa ra số chẵn.
c) d) tương tự
bài làm
62x+3.2x=9.29
62x+6x=261
62+6x=261
68x=261
x=261:68
x=\(\frac{261}{68}\)