1. Tìm x :
a, 4x2 - 9 = 0
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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
B
1
NH
21 tháng 6 2023
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
PT
1
CM
3 tháng 6 2017
Điều kiện để A xác định: x 2 – 10 x + 9 ≠ 0 ⇔ (x - 1)(x - 9) ≠ 0 ⇔ x ≠ 1, x ≠ 9
Ta có:
Để A = 0 ⇔ 
⇔ x - 4 = 0 ⇒ x = 4(tm đk)
Vậy với x = 4 thì A = 0
C
2
7 tháng 6 2021
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
7 tháng 6 2021
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Y
21 tháng 6 2023
\(a,x^2=5\Leftrightarrow x=\pm\sqrt{5}\)
Vậy \(S=\left\{\pm\sqrt{5}\right\}\)
\(b,3x^2-12=0\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
Vậy \(S=\left\{\pm2\right\}\)
\(c,4x^2-3=-9\)
\(\Leftrightarrow4x^2=-6\)
\(\Leftrightarrow x^2=-\dfrac{3}{2}\) (loại)
Vậy pt vô nghiệm.
\(d,5x^2-3=-3\)
\(\Leftrightarrow5x^2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
NH
21 tháng 6 2023
a)
`x^2 =5`
`=>\(\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
b)
`3x^2 -12=0`
`<=>3x^2 =12`
`<=>x^2 =4`
\(< =>\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
c)
`4x^2 -3=-9`
`<=>4x^2 =-6`
`<=>x^2 =-3/2` (vô lí vì `x>=0AA x` )
d)
`5x^2 -3=3`
`<=>5x^2 =0`
`<=>x^2 =0`
`<=>x=0`
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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
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: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
10 tháng 7 2021
Lời giải:
$4x^2-2x-1=0$
$\Leftrightarrow [(2x)^2-2.2x.\frac{1}{2}+(\frac{1}{2})^2]-\frac{5}{4}=0$
$\Leftrightarrow (2x-\frac{1}{2})^2=\frac{5}{4}$
$\Rightarrow 2x-\frac{1}{2}=\pm \frac{\sqrt{5}}{2}$
$\Leftrightarrow 2x=\frac{1\pm \sqrt{5}}{2}$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{4}$
$x^4-4x^2-32=0$
$\Leftrightarrow (x^2-2)^2-36=0$
$\Leftrightarrow (x^2-2-6)(x^2-2+6)=0$
$\Leftrightarrow (x^2-8)(x^2+4)=0$
Vì $x^2+4>0$ với mọi $x$ nên $x^2-8=0$
$\Leftrightarrow x=\pm 2\sqrt{2}$
10 tháng 7 2021
a) Ta có: \(4x^2-2x-1=0\)
\(\Delta=\left(-2\right)^2-4\cdot4\cdot\left(-1\right)=4+16=20\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2-2\sqrt{5}}{8}=\dfrac{1-\sqrt{5}}{4}\\x_2=\dfrac{2+2\sqrt{5}}{8}=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
b) Ta có: \(x^4-4x^2-32=0\)
\(\Leftrightarrow x^4-8x^2+4x^2-32=0\)
\(\Leftrightarrow x^2=8\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
\(4x^2-9=0\)
\(4x^2=9\)
\(x^2=\frac{9}{4}\)
\(x^2=\frac{3}{2}^2\)
\(x=\frac{3}{2}\)
4x2 - 9 = 0
4x2 = 0 + 9
4x2 = 9
( 2x )2 = 32
=> 2x = 3
x = \(\frac{3}{2}\)