Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\dfrac{2}{1+2x}+\dfrac{4x^2+1}{4x^2-1}-\dfrac{1}{1-2x}\right):\dfrac{2}{4x^2-1}\)
\(=\left(\dfrac{2\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}+\dfrac{-\left(4x^2+1\right)}{\left(1-2x\right)\left(1+2x\right)}-\dfrac{1\left(1+2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{4x^2-1}{2}\)
\(=\left(\dfrac{2-4x-4x^2-1-1-2x}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{\left(1-2x\right)\left(1+2x\right)}{-2}\)
\(=\left(\dfrac{-4x^2-6x}{\left(1+2x\right)\left(1-2x\right)}\right)\cdot\dfrac{\left(1-2x\right)\left(1+2x\right)}{-2}\)
\(=\dfrac{-2x\left(2x+3\right)\left(1-2x\right)\left(1+2x\right)}{\left(1+2x\right)\left(1-2x\right)\cdot\left(-2\right)}\)
\(=\dfrac{x\left(2x+3\right)}{1}\)
\(=x\left(2x+3\right)\)
Để A = 2 thì \(x\left(2x+3\right)=2=1\cdot2=2\cdot1=\left(-1\right)\cdot\left(-2\right)=\left(-2\right)\cdot\left(-1\right)\)
Ta có bảng :
| x | 1 | 2 | -1 | -2 |
| 2x+3 | 2 | 1 | -2 | -1 |
| x1 | 1 | 2 | -1 | -2 |
| x2 | -0,5 | -1 | -2,5 | -2 |
Ta thấy chỉ có x = -2 và 2x + 3 = -1 thì x1 và x2 mới bằng nhau và bằng -2
Vậy x = -2 thì A = 2
`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`
`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`
`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`
`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`
`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`
`=(2x+6)/(x-2)`
ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)
\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)
\(=\dfrac{\left(3+x\right)^2-\left(3-x\right)^2-4x^2}{\left(3+x\right)\left(3-x\right)}:\dfrac{x-2}{x+3}\)
\(=\dfrac{\left(3+x-3+x\right)\left(3+x+3-x\right)-4x^2}{\left(x+3\right)\left(3-x\right)}.\dfrac{x+3}{x-2}\)
\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)
\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)
1) \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
2) \(\left(2x-3\right)^2=4x^2-12x+9\)
\(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(2x-5\right)^2=4x^2-20x+25\)
\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)
3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)
\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)
\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}\)
\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)
1: \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
\(\dfrac{x^2+x+1}{x^2-x+1}-\dfrac{1}{3}=\dfrac{3x^2+3x+3-x^2+x-1}{3\left(x^2-x+1\right)}\)
\(=\dfrac{2x^2+4x+2}{3\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge0\)
Do đó: \(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\)(1)
\(\dfrac{x^2+x+1}{x^2-x+1}-3=\dfrac{x^2+x+1-3x^2+3x-3}{x^2-x+1}\)
\(=\dfrac{-2x^2+4x-2}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)
Do đó: \(\dfrac{x^2+x+1}{x^2-x+1}\le3\)(2)
Từ (1)và (2) suy ra ĐPCM
\(a,x\left(-3x+5\right)+3x\left(x+1\right)-40=0\)
\(\left(x.-3x\right)+\left(5x\right)+3x\left(x+1\right)-40=0\)
\(-3x^2+5x+\left(3x.x\right)+\left(3x.1\right)-40=0\)
\(-3x^2+5x+3x^2+3x-40=0\)
\(\left(-3x^2+3x^2\right)+5x+3x-40=0\)
\(8x-40=0\)
\(8x=0+40=40\)
\(x=40:8=5\)
a) \(x\left(5-3x\right)+3x\left(x+1\right)-40=0\)
\(\Rightarrow5x-3x^2+3x^2+3x-40=0\)
\(\Rightarrow8x-40=0\)
\(\Rightarrow8x=40\)
\(\Rightarrow x=5\)
b) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
\(\Rightarrow83x=83\)
\(\Rightarrow x=1\)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
`2+1/[2+1/[2+1/[2+1/2]]]`
`=2+1/[2+1/[2+1/[4/2+1/2]]]`
`=2+1/[2+1/[2+1/[5/2]]]`
`=2+1/[2+1/[2+2/5]]`
`=2+1/[2+1/[10/5+2/5]]`
`=2+1/[2+1/[12/5]]`
`=2+1/[2+5/12]`
`=2+1/[24/12+5/12]`
`=2+1/[29/12]`
`=2+12/29`
`=58/29+12/29=70/29`
\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{\dfrac{5}{2}}}}=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{2}{5}}}=2+\dfrac{1}{2+\dfrac{1}{\dfrac{12}{5}}}=2+\dfrac{1}{2+\dfrac{5}{12}}=2+\dfrac{1}{\dfrac{29}{12}}=2+\dfrac{12}{29}=\dfrac{70}{29}\)