Tìm x:
(5x^4 - 3x^3):2x^3=1/2
M.n giúp hộ vs, mk đag cần gấp
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a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
với |2x+10|+|3x-1|+|1-x|=3 ta có 2 trường hợp:
trường hợp 1:|2x+10|+|3x-1|+|1-x|=2x+10+3x-1+1-x=3
4x+10=3
4x=-7
x=-7/4
trường hợp 2:|2x+10|+|3x-1|+|1-x|=-(2x+10)+[-(3x-1)]+[-(1-x)]=3
-2x-10-3x+1-1+x=3
-4x-10=3
-4x=13
x=-13/4
/ là dấu phần nhé!
a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84
<=> -80x = -480
<=> x = 6
b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1
<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0
<=> -4x = 0
<=> x = 0
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
= 10x - 16 - 12x + 15 = 12x - 16 + 11
= -14x = -4
= x =\(\frac{2}{7}\)
d) 5x-3{4x-2[4x-3(5x-2)]}=182
= 5x - 3 . [4x - 2(4x - 15x + 6)]
= 5x - 3 . (4x - 8x + 30x - 12)
= 5x - 12x + 24x - 90x + 36
= -73x + 36 = 182
=> -73x = 182 - 36 = 146
=> x = 146 : (-73) = -2
~Hok tốt~
a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)
Vậy \(x=\frac{-3}{4}\)
b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
Điều kiện x khác 0
\(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\)
\(\Rightarrow\frac{5}{2}x-\frac{3}{2}=\frac{1}{2}\)
\(\Rightarrow\frac{5}{2}x=2\Rightarrow x=\frac{4}{5}\)