(2x+3)2 =9 phần 121
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DT
3
9 tháng 8 2016
(2x + 3)2 = 9/121
=> 2x + 3 thuộc {3/11 ; -3/11}
=> 2x thuộc {-30/11 ; -36/11}
=> x thuộc {-15/11 ; -18/11}
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^^
BV
a) (x^4)^2=x^12/x^5 (x khác 0 )
b) x^10+=25x^8
c) (2x+3)^2=9/121
d) (3x-1)^3=-8/27
Các bạn giúp mình với
0
NH
1
TN
24 tháng 7 2018
\(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x-\frac{1}{2}=0\)
<=> \(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
<=> \(\hept{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(\left(2x+3\right)^2=\frac{9}{121}\)
<=-> \(\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)
<=> \(\hept{\begin{cases}2x=\frac{-30}{11}\\2x=\frac{-36}{11}\end{cases}}\)
\(2x^{10}=25x^8\)
<=> \(2x^{10}-25x^8=0\)
<=> \(x^8.\left(2x^2-25\right)=0\)
<=> \(\hept{\begin{cases}x^8=0\\2x^2-25=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x^2=\frac{25}{2}\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{25}{2}}\\x=-\sqrt{\frac{25}{2}}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{cases}}\)
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 ^^
A
2
NB
30 tháng 9 2019
a)\(\left(2x+3\right)^2=\frac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{matrix}\right.\)
Vậy...
b)\(\left(3x-1\right)^3=\frac{-8}{27}\\ \Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\\ 3x-1=\frac{-2}{3}\\ \Rightarrow x=\frac{1}{9}\)
Vậy...
VM
30 tháng 9 2019
a) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow2x+3=\pm\frac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{3}{11}-3=-\frac{30}{11}\\2x=\left(-\frac{3}{11}\right)-3=-\frac{36}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-\frac{30}{11}\right):2\\x=\left(-\frac{36}{11}\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}.\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(\Rightarrow3x=\left(-\frac{2}{3}\right)+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
Chúc bạn học tốt!
(2x+3)2 \(=\) \(\dfrac{9}{121}\)
(2x+3)2 \(\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=\dfrac{-3}{11}\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}2x=\dfrac{3}{11}-3\\2x=\left(\dfrac{-3}{11}-3\right)\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}2x=\dfrac{3}{11}-\dfrac{33}{11}\\2x=\left(\dfrac{-3}{11}-\dfrac{33}{11}\right)\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}2x=\dfrac{-30}{11}\\2x=\dfrac{-36}{11}\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{-30}{11}:2\\x=\dfrac{-36}{11}:2\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{-15}{11}\\x=\dfrac{-18}{11}\end{matrix}\right.\)
Vậy x \(=\) \(\dfrac{-15}{11}\) và \(\dfrac{-18}{11}\)
Ý cậu là vậy hả ?
\(\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Leftrightarrow2x+3=\dfrac{3}{11}\)
\(\Leftrightarrow2x=\dfrac{-30}{11}\)
\(\Leftrightarrow x=\dfrac{-15}{11}\)
Vậy \(x=\dfrac{-15}{11}\)