tìm \(x\in Z\)biết :
\(a,2x-138=3^2\cdot2^3\)
\(b,6x-39=588\div28\)
\(c,42\cdot x+37\cdot42=39\cdot42\)
\(d,71+\left(26-3x\right)\div5=75\)
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a)\(2x-138=3^2.2^3\)
\(2x-138=72\)
\(x=105\)
b)\(x=2^4+3^2.3^2\)
\(x=16+81\\ x=97\)
c)\(6x-39=\frac{588}{28}\\ 6x-39=21\\ x=10\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.42}\)
\(=\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3.5^3.2^4.2.3.7}=\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3^2.5^3.2^5.7}=\frac{-2.3}{1}=-6\)
học tốt~~~
A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
a) 2x - 138 = 32 . 23
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 105
b) 6x - 39 = 588 : 28
6x - 39 = 21
6x = 21 + 39
6x = 60
x = 60 : 6 = 10
c)42 . x + 37 . 42 = 39 . 42
42 . (x + 37) = 1638
x + 37 = 1638 : 42
x + 37 = 39
x = 39 - 37
x = 2