$(2x+\dfrac 3 5)^2-\dfrac{24}{25}=1\\\Leftrightarrow (2x+\dfrac{3}{5})^2=\dfrac{49}{25}\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{3}{5}=\dfrac{7}{5}\\2x+\dfrac{3}{5}=-\dfrac{7}{5}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}2x=\dfrac{4}{5}\\2x=-2\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{2}{5}\\x=-1\end{array}\right.$

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 \(x=\frac{-2}{5}\)

-2/5

chắc chắn

\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)

\(\left(4-8x\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)

\(2\left(x-4\right)-3x\left(x-4\right)=0\)

\(\left(2-3x\right)\left(x-4\right)=0\)

\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

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

330 nhé 

tk mình nha  Nguyễn Thanh Huyền

Tui cux k bt nữa tui tính rồi mà tinhs mãi k ra 

Bà bt ln chưa chỉ tôi vs

Mọi người ơi giúp mk đi

(2x+3)\(^2\) = \(\frac{25}{9}\)

=> 2x+3 = \(\frac{5}{3}\)

=> 2x = \(\frac{5}{3}\) - 3

=> 2x = \(-\frac{4}{3}\)

=> x =\(-\frac{2}{3}\)

tích cho mk trước đi

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

Mk đang tắc chỗ đó đấy

Dù sao cũng Thanks bạn nhé