Hộ cái
Jup nha mn nhanh gọn lẹ đg cần gấp
Tinm x E Z,bt:
a) x-14=3x+18 ;è)(x-5)-3(x-4) =-6+15(-3)
b)(x+7)(x-9)=0 ;g)|2x-5|-7=22
c)x(x+3)=0 ; h)-12(x-5)+7(3-x=5
d)(x-2)(5-x)=0 ;i)30(x+2)-6(x-5)-24x=100
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= 144 . x = 56 + 130. x
= 144 . x - 130 . x = 56
= x.[144 - 130] = 56
= x.14 = 56
= x = 56 : 14
= x = 4
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{81}{256}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x-1}=\left(\frac{-3}{4}\right)^4\)
\(\Rightarrow3x-1=4\)
\(\Rightarrow3x=4+1\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=5:3\)
\(\Rightarrow x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
_Chúc bạn học tốt_
a) \(\left(3x+4\right)^2-64=0\)
\(\left(3x+4\right)^2=0+64\)
\(\left(3x+4\right)^2=64\)
=> \(3x+4=8\) hay \(3x+4=-8\)
\(3x=8-4\) hay \(3x=-8-4\)
\(3x=4\) hay \(3x=-12\)
\(x=4:3\) hay \(x=-12:3\)
\(x=\dfrac{4}{3}\) hay \(x=-4\)
Vậy \(x=\dfrac{4}{3}\)hoặc \(x=-4\)
3x - 18 = 15 - x
=> 3x - x = 18 + 15
=> 4x - x = 33
=> x = 33 / 4 = 8,25 .
3x - 18 = 15 - x
=> 3x + x = 15 + 18
=> 4x = 33
=> x = 33/4 = 8,25
ta có: f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)
\(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)
\(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)
\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)
Chúc bn học tốt !!!!!!
Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............
a) x-14=3x + 18
x - 3x = 18 + 14
-2x = 32
=> x = -16
b) (x+7)(x-9)=0
=> TH1: x+7=0 => x = -7
=> TH2: x-9=0 => x = 9
c) x(x+3) =0
=> TH1: x=0
=> TH2: x+3 =0 => x = -3
d) (x-2)(5-x)=0
=> TH1: x-2=0 => x=2
=> Th2: 5-x=0 => x=5