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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: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
K
1 tháng 12 2019
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow-2x-7=2x+5\)
\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right].\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
9 tháng 8 2021
a)4x2-9=0
⇔ (2x-3)(2x+3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b)(x+5)2-(x-1)2=0
⇔ (x+5-x+1)(x+5+x-1)=0
⇔ 12(x+2)=0
⇔ x=-2
c)x2-6x-7=0
⇔ x2-7x+x-7=0
⇔ x(x-7)+(x-7)=0
⇔ (x-7)(x+1)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
d)(x+1)2-(2x-1)2=0
⇔ (x+1-2x+1)(x+1+2x-1)=0
⇔3x(2-x)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
HN
9 tháng 8 2021
a, 4x2 - 9 = 0
<=> 4x2 = 9
<=> x2 = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)
b, (x + 5 )2 - ( x - 1 )2 = 0
<=> ( x+5-x+1 )(x+5+x-1) = 0
<=> 6(2x+4) = 0
<=> 12x+24=0
<=> 12x = -24
<=> x = -2
c, x2-6x-7=0
<=> x2+x-7x-7=0
<=> x(x+1)-7(x+1)=0
<=> (x-7)(x+1)=0
=> x+7=0 hoặc x+1=0
+ x-7=0 => x=7
+ x+1=0 => x=-1
d, \(\left(x+1\right)^2-\left(2x-1\right)^2=0\)
<=> \(\left(x+1-2x+1\right)\left(x+1+2x-1\right)=0\)
<=> (-x+2).3x=0
=> x=0 hoặc (-x+2).3=0
+ (-x+2).3=0 => -3x+6=0 => x=-2
CM
6 tháng 7 2018
( 2 x – 3 ) 2 – 4 x 2 + 9 = 0 ⇔ ( 2 x – 3 ) 2 – ( 4 x 2 – 9 ) = 0 ⇔ ( 2 x – 3 ) 2 – ( ( 2 x ) 2 – 3 2 ) = 0 ⇔ ( 2 x – 3 ) 2 – ( 2 x – 3 ) ( 2 x + 3 ) = 0
ó (2x – 3)(2x – 3 – 2x – 3) = 0
ó (2x – 3)(-6) = 0
ó 2x – 3 = 0
ó x = 3 2
Đáp án cần chọn là: C
C
3
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
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) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\Rightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
HV
2
11 tháng 10 2021
1: Ta có: \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
LN
2
7 tháng 8 2021
1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)
\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)
2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)
3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)
7 tháng 8 2021
1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
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
\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)
\(\Rightarrow\left[\left(2x+5\right)\left(2x-5\right)\right]^2-9\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)^2\left[\left(2x+5\right)^2-3^2\right]=0\)
\(\Rightarrow\left(2x-5\right)^2\left(2x+5-3\right)\left(2x+5+3\right)=0\)
\(\Rightarrow\left(2x-5\right)^2=0\Rightarrow2x-5=0\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
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