Tìm số nguyên x
\(9^x:3^x=3\)
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a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
a: \(P=\dfrac{x}{x+3}-\dfrac{x^2-5x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{3}{x-3}\)
\(=\dfrac{x^2-3x-x^2+5x+6+3x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{5}{x-3}\)
Bài 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
x − 3 y − 3 = 9 ⇔ x − 3 = ± 3 y − 3 = ± 3 x − 3 = ± 1 y − 3 = ± 9 x − 3 = ± 9 y − 3 = ± 1 TH 1 : x − 3 = 3 y − 3 = 3 ⇔ x = 3 + 3 y = 3 + 3 ⇔ x = 6 y = 6 TH 2 : x − 3 = − 3 y − 3 = − 3 ⇔ x = − 3 + 3 y = − 3 + 3 ⇔ x = 0 y = 0 TH 3 : x − 3 = 9 y − 3 = 1 ⇔ x = 9 + 3 y = 1 + 3 ⇔ x = 12 y = 4 TH 4 : x − 3 = − 9 y − 3 = − 1 ⇔ x = − 9 + 3 y = − 1 + 3 ⇔ x = − 6 y = 2 TH 5 : x − 3 = 1 y − 3 = 9 ⇔ x = 1 + 3 y = 9 + 3 ⇔ x = 4 y = 12 TH 6 : x − 3 = − 1 y − 3 = − 9 ⇔ x = − 1 + 3 y = − 9 + 3 ⇔ x = 2 y = − 6 x ; y ∈ 2 ; − 6 ; 6 ; 6 ; 0 ; 0 ; 12 ; 4 ; − 6 ; 2 ; 4 ; 12
x − 3 y − 3 = 9 ⇔ x − 3 = ± 3 y − 3 = ± 3 x − 3 = ± 1 y − 3 = ± 9 x − 3 = ± 9 y − 3 = ± 1
T H 1 : x − 3 = 3 y − 3 = 3 ⇔ x = 3 + 3 y = 3 + 3 ⇔ x = 6 y = 6
T H 2 : x − 3 = − 3 y − 3 = − 3 ⇔ x = − 3 + 3 y = − 3 + 3 ⇔ x = 0 y = 0
T H 3 : x − 3 = 9 y − 3 = 1 ⇔ x = 9 + 3 y = 1 + 3 ⇔ x = 12 y = 4
T H 4 : x − 3 = − 9 y − 3 = − 1 ⇔ x = − 9 + 3 y = − 1 + 3 ⇔ x = − 6 y = 2
T H 5 : x − 3 = 1 y − 3 = 9 ⇔ x = 1 + 3 y = 9 + 3 ⇔ x = 4 y = 12
T H 6 : x − 3 = − 1 y − 3 = − 9 ⇔ x = − 1 + 3 y = − 9 + 3 ⇔ x = 2 y = − 6
Vậy x ; y ∈ 2 ; − 6 ; 6 ; 6 ; 0 ; 0 ; 12 ; 4 ; − 6 ; 2 ; 4 ; 12
Đáp án là C
3|x + 1| = 9 ⇒ |x + 1| = 3
⇒ x + 1 = 3 hay x = 2
Hoặc x + 1 = -3 hay x = -4
\(9^x:3^{x=3}\)
\(\left(9:3\right)^x=3\)
\(3^x=3\)
\(\Rightarrow x=1\)
=> (9:3)x=31
=> 3x=31
=> x=1