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ta có : \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
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x/-15=-60/x
x2012=2012
(3x-1)3=-8/27
-2/x=-x/8/25
(8x-1)2n=52n
(x-2/3)3=(2/3)6
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\(\left(3x-1\right)^3=\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{2}{3}\)
\(\Rightarrow3x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:3\)
\(\Rightarrow x=\frac{5}{9}\)
tíc mình nha
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\(a/\left(3x+\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3\Leftrightarrow3x+\frac{1}{2}=-\frac{3}{2}\Leftrightarrow x=-\frac{2}{3}\)
\(b/-3x^2+15=0\Leftrightarrow3\left(5-x^2\right)=0\Leftrightarrow5-x^2=0\Leftrightarrow x=\pm\sqrt{5}\)
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a) (2x + 3)2 = \(\frac{9}{121}=\left(\frac{3}{11}\right)^2=\left(-\frac{3}{11}\right)^2\)
Trường hợp 1: \(2x+3=\frac{3}{11}\)
\(2x=\frac{3}{11}-3=-\frac{30}{11}\)
\(x=-\frac{30}{11}:2=-\frac{15}{11}\)
Trường hợp 2: \(2x+3=-\frac{3}{11}\)
\(2x=-\frac{3}{11}-3=-\frac{36}{11}\)
\(x=-\frac{36}{11}:2=-\frac{18}{11}\)
Vậy \(x=-\frac{15}{11}\)hoặc \(x=-\frac{18}{11}\)
b,(3x-1)3= -8/27= (-2/3)^3
<=> 3x-1 = =2/3
<=>x=1/9 Mjk thấy phần a có bạn lm rồi nên bổ sung phần b
Chúc các bạn học tốt nhé^^
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1) \(3^{x+1}=27\)
\(\Leftrightarrow3^{x+1}=3^3\)
\(\Leftrightarrow x+1=3\Leftrightarrow x=2\)
2) \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=3\Leftrightarrow2x=4\Leftrightarrow x=2\)
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(3x - 1)3 = -8/27
=> (3x - 1)3 = (-2/3)3
=> 3x - 1 = -2/3
=> 3x = -2/3 + 1
=> 3x = 1/3
=> x = 1/3 : 3
=> x = 1/9
x3 : 3 = 9
=> x3 = 9.3
=> x3 = 27
=> x3 = 33
=> x = 3
x10 = 25.x8
=> x10 : x8 = 25
=> x10-8 = 25
=> x2 = 52 = (-5)2
=> x = 5 hoặc x = -5
Vậy x \(\in\){-5; 5}.
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a) \(\left(2x+3\right)^2=\frac{9}{121}\)
Ta có: \(\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)
\(\Rightarrow2x+3\in\left\{\frac{3}{11};\frac{-3}{11}\right\}\)
\(\Rightarrow x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)
Vậy \(x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)
b) \(\left(3x-1\right)^3=\frac{-8}{27}\)
Ta có: \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
a.
\(\left(2x+3\right)^2=\frac{9}{121}\)
\(\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\)
\(2x+3=\pm\frac{3}{11}\)
TH1:
\(2x+3=\frac{3}{11}\)
\(2x=\frac{3}{11}-3\)
\(2x=-\frac{30}{11}\)
\(x=-\frac{30}{11}\div2\)
\(x=-\frac{15}{11}\)
TH2:
\(2x+3=-\frac{3}{11}\)
\(2x=-\frac{3}{11}-3\)
\(2x=-\frac{36}{11}\)
\(x=-\frac{36}{11}\div2\)
\(x=-\frac{18}{11}\)
Vậy \(x=-\frac{15}{11}\) hoặc \(x=-\frac{18}{11}\)
b.
\(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(3x-1=-\frac{2}{3}\)
\(3x=-\frac{2}{3}+1\)
\(3x=\frac{1}{3}\)
\(x=\frac{1}{3}\div3\)
\(x=\frac{1}{9}\)
Chúc bạn học tốt ^^
\(\left(3x-1\right)^3=-\dfrac{8}{27}\\ \Leftrightarrow3x-1=-\dfrac{2}{3}\\ \Leftrightarrow3x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
\(\Leftrightarrow3x-1=-\dfrac{2}{3}\)
hay x=1/9