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S
2
16 tháng 11 2021
\(\left(5x-2\right)\left(2x+7\right)-4x^2-25=0\)
\(10x+35-4x^2-14x-4x^2+25=0\)
\(-4x+60-8x^2=0\)
\(-4\left(2x^2+x-15\right)=0\)
\(-4\left(2x^2+6x-5x-15\right)=0\)
\(-4\left(2x-5\right)\left(x+3\right)=0\)
=> \(x\) ∈ \(\left\{\dfrac{5}{2};-3\right\}\)
BT
1
DT
5 tháng 8 2016
1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right).-2=0\)
\(\Leftrightarrow-4x+10=0\)
\(\Leftrightarrow-4x=-10\)
\(\Leftrightarrow x=\frac{5}{2}.\)
Vậy \(S=\left\{\frac{5}{2}\right\}\)
2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\)
\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)
\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)
Vậy \(S=\left\{-3;0;2\right\}\)
18 tháng 8 2018
\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2=0\)
\(\left[\left(2x-5\right)\left(2x+5\right)\right]^2-9\left(2x-5\right)^2=0\)
\(\left(2x-5\right)^2\left[\left(2x+5\right)^2-9\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-5\right)^2=0\\\left(2x+5\right)-9=0\end{cases}}\)
+) \(\left(2x-5\right)^2=0\)
\(\Rightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
+) \(\left(2x+5\right)^2-9=0\)
\(\Leftrightarrow\left(2x+5-3\right)\left(2x+5+3\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+2=0\\2x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
Vậy ....
NT
4
D
21 tháng 10 2016
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
NV
18 tháng 8 2015
\(\Rightarrow\left[\left(2x+5\right)\left(2x-5\right)\right]^2-9\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)^2\left[\left(2x+5\right)^2-3^2\right]=0\)
\(\Rightarrow\left(2x-5\right)^2\left(2x+5-3\right)\left(2x+5+3\right)=0\)
\(\Rightarrow\left(2x-5\right)^2=0\Rightarrow2x-5=0\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
hoặc \(2x+2=0\Rightarrow2x=-2\Rightarrow x=-1\)
hoặc \(2x+8=0\Rightarrow2x=-8\Rightarrow x=-4\)
vậy x = 5/2 ; x = -1 ; x = -4
T
22 tháng 10 2023
\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(---\)
\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(---\)
\(c,4x(x-2)-x(3+4x)(?)\)
\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
\(---\)
\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(---\)
\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(Toru\)
ĐKXĐ:\(2x-5\ne0\Rightarrow x\ne\dfrac{5}{2}\)
\(\dfrac{4x^2-25}{2x-5}=0\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(2x+5\right)}{2x-5}=0\\ \Leftrightarrow2x+5=0\\ \Leftrightarrow x=-\dfrac{5}{2}\left(tm\right)\)
\(\dfrac{4x^2-25}{2x-5}\)
\(=\dfrac{\left(2x-5\right)\left(2x+5\right)}{2x-5}\)\(=2x+5=0\)
\(\Rightarrow x=-\dfrac{5}{2}\)