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a)\(\frac{20^3}{40^3}=\frac{1^3}{2^3}=\frac{1}{8}\)
b) \(\frac{390^4}{130^4}=\frac{3^4}{1^4}=\frac{81}{1}=81\)
c) \(\frac{3^2}{0,375^2}=\frac{8^2}{1^2}=\frac{64}{1}=64\)
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Bài 2: Mình nghĩ câu a là a+2b-3c=-20
a) Ta có: a/2 = b/3 = c/4 = 2b/6 = 3c/12 = a + 2b - 3c/ 2 + 6 - 12 = -20/-4 = 5
a/2 = 5 => a = 2 . 5 = 10
b/3 = 5 => b = 5 . 3 = 15
c/4 = 5 => c = 5 . 4 = 20
Vậy a = 10; b = 15; c = 20
b) Ta có: a/2 = b/3 => a/10 = b/15
b/5 = c/4 => b/15 = c/12
=> a/10 = b/15 = c/12 = a - b + c / 10 - 15 + 12 = -49/7 = -7
a/10 = -7 => a = -7 . 10 = -70
b/15 = -7 => b = -7 . 15 = -105
c/12 = -7 => c = -7 . 12 = -84
Vậy a = -70; b = -105; c = -84.
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\(4+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=4+\frac{1}{1+\frac{1}{1+\frac{2}{\frac{7}{4}}}}=4+\frac{1}{1+\frac{1}{1+\frac{8}{7}}}=4+\frac{1}{1+\frac{1}{\frac{15}{7}}}=4+\frac{1}{1+\frac{7}{15}}=4+\frac{1}{\frac{22}{15}}=4+\frac{15}{22}=\frac{103}{22}\)
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a, 2/1/3:1/3=7/9:x b, x:1/3=12/99:15/90
7 = 7/9 : x x :1/3= 8/11
x = 7/9:7 x = 8/11 * 1/3
x = 1/9 x = 8/33
c, 0,15:x=3/1/3:2,25 d, 3/4:0,75=x:75/90
0,15:x =40/27 1 =x:75/90
x = 0,15:40/27 x = 1*75/90
x = 81/800 x = 75/90
e, x/-15=-60/x f, -2/x=-x/8
=>x*x=-15*(-60) => (-2)*8=x*-x
=>x2=900 => -16 = -x2
=>x2=302 hoặc x2=(-30)2 => 16=x2
=> x=30 hoặc x= -30 => x2=42 hoặc x2=(-4)2
=> x=4 hoặc x=-4
a)\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(\frac{7}{3}\times3=\frac{7}{9}:x\)
\(7=\frac{7}{9}:x\)
\(x=\frac{7}{9}:7\)
\(x=\frac{7}{9}\times\frac{1}{7}\)
\(x=\frac{1}{9}\)
\(\frac{390^4}{90^3}=\frac{30^4.13^4}{30^3.3^3}=\frac{30.13^4}{3^3}=\frac{10.13^4}{9}\)
\(\frac{390^4}{90^3}=\frac{13^4.30^4}{3^3.30^3}=\frac{13^4.30}{27}\)