\(\left(2x+3\right)^2=\frac{9}{121}\\ \left(3x-1\right)^3=-\frac{8}{27}\)
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SU
2
3 tháng 8 2015
a, \(\left(2x+3\right)^2=\frac{3^2}{11^2}\)
từ đó suy ra
\(2x+3=\frac{3}{11}\)
2x=3/11-3
2x=-2/8/11
x=-2/8/11:2
x=-1/4/11
b,
(3x-1)^3=-8/27
(3x-1)^3=(-2/3)^3
Vậy suy ra
3x-1=-2/3
3x=-2/3+1
3x=1/3
x=1/3:3
x=1/9
VM
14 tháng 2 2020
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
MS
6 tháng 6 2016
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
6 tháng 6 2016
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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...
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!