4x2 - 25 - (2x - 5)( 2x + 7) = 0
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LM
1
29 tháng 10 2021
g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Rightarrow-2\left(2x-5\right)=0\Rightarrow x=\dfrac{5}{2}\)
i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\Rightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
K
1 tháng 12 2019
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow-2x-7=2x+5\)
\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right].\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
31 tháng 8 2021
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Rightarrow\left(2x-5\right)\left(-2x-7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Rightarrow-2x-7=2x+5\)
\(\Rightarrow4x=-12\)
\(\Rightarrow x=-3\)
20 tháng 7 2021
\(4x^2-3\left(2x-5\right)-25=0\Leftrightarrow4x^2-6x-10=0\)
\(\Leftrightarrow2\left(2x^2-3x-5\right)=0\Leftrightarrow2\left(x+1\right)\left(2x-5\right)=0\Leftrightarrow x=-1;x=\dfrac{5}{2}\)
12 tháng 7 2021
Ta có: \(\left(2x-5\right)\left(4x^2+10x+25\right)\left(2x+5\right)\left(4x^2-10x+25\right)-64x^6\)
\(=\left(8x^3-125\right)\left(8x^3+125\right)-64x^6\)
\(=64x^6-15625-64x^6\)
=-15625
18 tháng 8 2021
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
18 tháng 8 2021
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
M
1
30 tháng 10 2021
a) \(\Leftrightarrow4x^2-10x-4x^2-3x=19\\ \Leftrightarrow-13x=19\\ \Leftrightarrow x=-\dfrac{19}{13}\)
b) \(\Leftrightarrow\left(x-7\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{5}{3}\end{matrix}\right.\)
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
7 tháng 11 2021
b) x(x-4) - 2x+8 = 0
x(x-4) - 2(x-4) = 0
(x-2) (x-4) = 0
TH1: x-2=0 TH2: x-4=0
x=2 x=4
Vậy x\(\in\){2;4}
7 tháng 11 2021
\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)
\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow4x^2-25-\left(4x^2+14x-10x-35\right)=0\\ \Leftrightarrow4x^2-25-4x^2-14x+10x+35=0\\ \Leftrightarrow-4x+10=0\)
\(\Leftrightarrow x=\frac{-10}{-4}\\ \Leftrightarrow x=\frac{5}{2}\)