3(x-4).(6x)=14
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a) \((4{x^3}):( - 2{x^2})\\= [4: (- 2)].({x^3}:{x^2})\\ = - 2.{x^{3 - 2}}\\ = - 2x\);
b) \(( - 7{x^2}):(6x) \\= ( - 7:6).({x^2}:x) \\= - \dfrac{7}{6}.{x^{2 - 1}}\\ = - \dfrac{7}{6}.x\);
c) \(( - 14{x^4}):( - 8{x^3}) \\= ( - 14: - 8).({x^4}:{x^3})\\= \dfrac{7}{4}.{x^{4 - 3}} \\= \dfrac{7}{4}.x\).
a)
x 8 = − 1 4 x .4 = − 1.8 x .4 = − 8 x = − 2
b)
6 x − 3 = 9 2 x − 7 6. 2 x − 7 = 9. x − 3 12 x − 42 = 9 x − 27 3 x = 15 x = 5
c)
x − 1 2 2 = 4 x − 1 2 = ± 2
TH1:
x − 1 2 = 2 x = 5 2
TH2:
x − 1 2 = − 2 x = − 3 2
Vậy x = 5 2 hoặc x = − 3 2
`x^4+6x^2-6x+14=0`
`<=>x^4+5x^2+6+x^2-6x+9=0`
`<=>x^4+5x^2+6+(x-3)^2=0`vô lý
Vì `x^4+5x^2+6+(x-3)^2>=6>0`
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
Bài 1:
a: \(x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
b: \(x^3-6x^2-9x+14\)
\(=x^3-7x^2+x^2-7x-2x+14\)
\(=\left(x-7\right)\left(x^2+x-2\right)\)
\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)
c: \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
Kiểm tra lại nhaa
\(6x\left(x-4\right)=\dfrac{14}{3}\Leftrightarrow6x^2-24x-\dfrac{14}{3}=0\Leftrightarrow x=\dfrac{6\pm\sqrt{43}}{3}\)