(2x+3/5)2-9/25=0
giup mk cai nha
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$(2x+\dfrac 3 5)^2-\dfrac{24}{25}=1\\\Leftrightarrow (2x+\dfrac{3}{5})^2=\dfrac{49}{25}\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{3}{5}=\dfrac{7}{5}\\2x+\dfrac{3}{5}=-\dfrac{7}{5}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}2x=\dfrac{4}{5}\\2x=-2\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{2}{5}\\x=-1\end{array}\right.$
Vậy $x=\dfrac{2}{5},x=-1$
GIải
\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{24}{25}=1\)
\(\left(2x+\dfrac{3}{5}\right)^2\) \(=1+\dfrac{24}{25}\)
\(\left(2x+\dfrac{3}{5}\right)^2\) \(=\dfrac{49}{25}\)
\(4x+\dfrac{9}{25}\) \(=\dfrac{49}{25}\)
\(4x\) \(=\dfrac{49}{25}-\dfrac{9}{25}\)
\(4x\) \(=\dfrac{8}{5}\)
\(x\) \(=4:\dfrac{8}{5}\)
\(x\) \(=\dfrac{5}{2}\)
\(\left(\frac{3}{5}\right)^{x+1}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{x+1}=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
\(\left(2x+1\right)^3=\frac{1}{125}\)
\(\left(2x+1\right)^3=\left(\frac{1}{5}\right)^3\)
\(\Rightarrow2x+1=\frac{1}{5}\)
\(\Rightarrow2x=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-2}{5}\)
vậy \(x=\frac{-2}{5}\)
\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)
\(\left(4-8x\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)
\(2\left(x-4\right)-3x\left(x-4\right)=0\)
\(\left(2-3x\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Ta thấy :
1 x 2 = 2
2 x 3 = 6
3 x 4 = 12
4 x 5 = 20
5 x 6 = 30
6 x 7 = 42
...
Khoảng cách giữa các tích là một dãy số 4 ; 6 ; 8 ; 10 ; 12 ; ...
Áp dụng công thức tính số số hạng ra kết quả 330
(2x+3)\(^2\) = \(\frac{25}{9}\)
=> 2x+3 = \(\frac{5}{3}\)
=> 2x = \(\frac{5}{3}\) - 3
=> 2x = \(-\frac{4}{3}\)
=> x =\(-\frac{2}{3}\)
TH2: (2x+3)\(^2\) =\(\frac{29}{5}\)
=> 2x+3 = \(-\frac{5}{3}\)
=> 2x = \(-\frac{5}{3}\) - 3
=> 2x = \(-\frac{14}{3}\)
=> x = \(-\frac{7}{3}\)
a) 4 + 9( x+6) = -2 + 25 = 23
=> 9(x+6) = 23 -4 =19
=> 9x + 54 = 19
=> 9x = 19 -54 =-35
=> x = \(-\dfrac{35}{9}\)
b) \(\dfrac{5}{x}-\dfrac{1}{8}=\dfrac{1}{2}+\left(-\dfrac{5}{12}\right)\)
\(\dfrac{5}{x}-\dfrac{1}{8}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{1}{12}\)
\(\dfrac{5}{x}=\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{5}{24}\)
\(\Rightarrow x=24\)
c) Đề đúng chứ ?
\(\left(3-x\right)^2+\frac{-9}{25}=\frac{2}{5}-\frac{8}{5}\)
\(\Rightarrow\left(3-x\right)^2+\frac{-9}{25}=\frac{-6}{5}\)
\(\Rightarrow\left(3-x\right)^2=\frac{-6}{5}+\frac{9}{25}\)
\(\Rightarrow\left(3-x\right)^2=\frac{-30}{25}+\frac{9}{25}\)
\(\Rightarrow\left(3-x\right)^2=\frac{-21}{25}\)
..đến đey thì có vấn đề =="
bạn ơi bài toán có sai đề bài ko nếu ko sai thì mình nghĩ lại
Sửa đề:
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\pm\frac{3}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=-\frac{3}{5}\)
Tham khảo nhé