(7392-X)÷4=79
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khi gap 1 thua so len 4 lan va giam thua so kia di 3 lan thi h moi la
7392*4/3 =9856
dap so :9856
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Thay \(a=7392\) ta có:
\(7392:83=89\left(dư5\right)\)
Vậy khi thay a = 7392 thì phép chia đó dư 5.
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4 x X + 79 x 2 = 79 x 33
4 x X + 158 = 2607
4 x X = 2607 - 158
4 x X = 2449
X = 2449 : 4
X = 612,25
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=>(1/2+1/3+...+1/80)*x>(1+1/79+1+2/78+...+1+78/2+1)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{80}\right)>\dfrac{80}{80}+\dfrac{80}{79}+...+\dfrac{80}{3}+\dfrac{80}{2}\)
=>x>80
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a)4*x+79*29=79*33
4*x+2291=2607
4*x=2607-2291=316
x=316:4=79
b)83-3*(7+x)=64
3*(7+x)=83- 64=19
7+x=19:3=19/3
x=19/3-7=-2/3
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a) Ta có:
\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)
\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)
\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)
Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.
Vậy A > C > B.
b) Ta có:
\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)
\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)
Vậy B : A = -4
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2 ^ x - 2 ^ 4 = 4 ^ 6 : 4 ^ 4
2 ^ x -16 = 4 ^ 2
2 ^ x - 16 = 16
2 ^ x = 16 + 16
2 ^ x = 32
2 ^ x = 2 ^ 5
suy ra x = 5
doi chut chua nghi ra phan b nhung nho nhe
(7392-x)÷4=79
(7392 - x) = 79 x 4
(7392 - x) ,= 316
x = 7392 - 316
x = 7076