Tìm giá trị biết
a) (2x+3) ²=9/121
b) (3x-1) ³=-8/27
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TN
2
22 tháng 7 2016
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 \(x=\frac{1}{9}\)
PA
22 tháng 7 2016
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 ^^
NT
2
23 tháng 7 2015
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 \(x=-\frac{18}{11}\)
20 tháng 9 2017
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^^
10 tháng 6 2017
a) Vì \(\left(2.x+3\right)^2=\dfrac{9}{121}\Rightarrow\left\{{}\begin{matrix}2.x+3=\dfrac{3}{11}\\2.x+3=-\dfrac{3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)
b) Vì \(\left(3.x-1\right)^3=-\dfrac{8}{27}\Rightarrow3.x-1=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{9}\)
HT
0
KM
2
KT
19 tháng 8 2018
a) \(\left(2x+3\right)^2=\frac{9}{21}\)
<=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\frac{4}{11}\\x=-1\frac{7}{11}\end{cases}}\)
Vậy...
K
31 tháng 8 2015
a) (2x + 3)2 = 9/121
Ta có: 9/121 = (3/11)2 = (-3/11)2
=> 2x + 3 thuộc {3/11; -3/11}
=> x thuộc {-15/11; -18/11}
b) (3x - 1)3 = -8/27 = (-2/3)3
=> 3x - 1 = -2/3
=> x = 1/9
7 tháng 12 2018
\(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\hept{\begin{cases}\left(\frac{3}{11}\right)^2\\\left(\frac{-3}{-11}\right)^2\end{cases}}\)
\(\Rightarrow2x+3=\hept{\begin{cases}\frac{3}{11}\\\frac{-3}{-11}\end{cases}}\)
a) (2x+3) ²=9/121
ta có \(\dfrac{9}{121}=\left(\dfrac{3}{11}\right)^2=\left(-\dfrac{3}{11}\right)^2\)
=> (2x+3) ∈ \(\left\{\dfrac{3}{11};\dfrac{-3}{11}\right\}\)
=> x ∈ \(\left\{\dfrac{-15}{11};-\dfrac{18}{11}\right\}\)
Vậy x ∈ \(\left\{\dfrac{-15}{11};-\dfrac{18}{11}\right\}\)
b) (3x-1) ³=-8/27
Ta có \(\dfrac{-8}{27}=\left(\dfrac{-2}{3}\right)^3\)
=> 3x-1 =\(\dfrac{-2}{3}\)
=> x = \(\dfrac{1}{9}\)
Vậy x = \(\dfrac{1}{9}\)
a, \(\left(2x+3\right)^2=\dfrac{9}{121}\Rightarrow\left(2x+3\right)^2=\left(\sqrt{\dfrac{9}{121}}\right)^2\)
\(\Rightarrow2x+3=\sqrt{\dfrac{9}{121}}=\pm\dfrac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=\dfrac{-3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\left(\dfrac{3}{11}-3\right)}{2}\\x=\dfrac{\left(\dfrac{-3}{11}-3\right)}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-15}{11}\\x=\dfrac{-18}{11}\end{matrix}\right.\)
b, \(\left(3x-1\right)^3=\dfrac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\dfrac{-8}{27}\Rightarrow\left(3x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\dfrac{-2}{3}\Rightarrow3x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{9}\)