(-3)3x+1=1/81
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2.(1 + 3 + 3² + ... + 3ˣ) + 1 = 81
2.(3ˣ⁺¹ - 1)/2 + 1 = 81
3ˣ⁺¹ - 1 + 1 = 81
3ˣ⁺¹ = 81
3ˣ⁺¹ = 3⁴
x + 1 = 4
x = 4 - 1
x = 3
Bài làm
a) 3x : 9 = 391
=> 3x : 32 = 391
=> 3x = 391 x 32
=> 3x = 393
=> x = 393 : 3
=> x = 392
Vậy x = 392
b) 81 : 3x + 1 = 9
81 : 3x = 9 - 1
81 : 3x = 8
3x = 81 : 8
3x = 10,125
x = 10,125 : 3
x = 3,375
Vậy x = 3,375
# Học tốt #
a ) \(3x:9=3^{91}\)
\(\Rightarrow3x:3^2=3^{91}\)
\(3x=3^{91}.3^2\)
\(3x=3^{93}\)
\(x=3^{93}:3\)
\(x=3^{92}\)
Vậy \(x=3^{92}\)
b ) \(81:3x+1=9\)
\(27x+1=9\)
\(27.x=9-1\)
\(27.x=8\)
\(x=8:27\)
\(x=\frac{8}{27}\)
Vậy \(x=\frac{8}{27}\)
Chúc bạn học tốt !!!
(12x - 5)(4x + 3) + (3x - 5)(1 - 16x) = 81
⇔ 48x2 + 36x - 20x - 15 + 3x - 48x2 - 5 + 80x - 81 = 0
⇔ 99x - 101 = 0
⇔ x = \(\dfrac{101}{99}\)
\(\left(\frac{-3}{4}\right)^{3x+5}=\left(\frac{81}{256}\right)^{-1}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x+5}=\frac{256}{81}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x+5}=\left(\frac{-3}{4}\right)^{-4}\)
\(\Rightarrow3x+5=-4\)
\(\Rightarrow3x=-9\)
\(\Rightarrow x=-3\)
Vậy x= -3
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)
\(=[(2x)^2-3^2](4x^2+9)\)
\(=(2x+3)(2x-3)(4x^2+9)\)
b, \(x^3-3x^2+3x-1=(x-1)^3\)
\(x^2-2x+1=(x-1)^2\)
c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)
\(=(6x^2+3x+1)(3x^3+1)\)
câu c bạn đánh sai 1 dấu phép toán kìa!!!!
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
a,
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)
\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
b,
( 3x + 1)^3 = 64
(3x + 1)^3 = 4^3
(3x + 1) = 4
3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
c,
( 2x - 3)^4 = 81
( 2x - 3) ^4 = 3^4
(2x - 3) = 3
2x = 3 + 3
2x = 6
x = 6: 2
x = 3