Giải các phương trình sau:a) 3x - 12 = 0b) (x-2)(2 x 3)=0c) x 2/x-2- 6/x 2= x^2/x^2-4Các bạn ơi giúp mình!!!!!

2

NT

Nguyễn Trọng Chiến

21 tháng 3 2021

a $\Leftrightarrow3x=12\Leftrightarrow x=4$

b $\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.$

c $ĐKXĐ:x\ne2;x\ne-2$

$\Rightarrow\left(x+2\right)^2-6\left(x-2\right)=x^2\Leftrightarrow x^2+4x+4-6x+12=x^2\Leftrightarrow-2x+16=0\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(TM\right)$

NL

Nguyễn Lê Phước Thịnh

21 tháng 3 2021

a) Ta có: 3x-12=0

$\Leftrightarrow3x=12$

hay x=4

Vậy: S={4}

b) Ta có: (x-2)(2x+3)=0

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.$

Vậy: $S=\left\{2;\dfrac{-3}{2}\right\}$

ST

see tình boi

12 tháng 1 2023

giải các phương trình sau:

a.(x - 1)(x + 2)= 0

b.(x -2)(x -5)=0

c.(x +3)(x -5)=0

d.(x + 1/2)(4x + 4)=0

e.(x -4)(5x -10)=0

f.(2x -1)(3x +6)=0

g.(2,3x -6,9)(0,1x -2)=0

4

TT

Thư Thư

12 tháng 1 2023

$a,\left(x-1\right)\left(x+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.$

$b,\left(x-2\right)\left(x-5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.$

$c,\left(x+3\right)\left(x-5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.$

$d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.$

$e,\left(x-4\right)\left(5x-10\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.$

$f,\left(2x-1\right)\left(3x+6\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.$

Đúng(4)

H

⭐Hannie⭐

12 tháng 1 2023

`a,(x-1)(x+2)=0`

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.$

`b,(x -2)(x -5)=0`

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.$

`c,(x +3)(x -5)=0`

$\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.$

`d,(x + 1/2)(4x + 4)=0`

$\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.$

`e,(x -4)(5x -10)=0`

$\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.$

`f,(2x -1)(3x +6)=0`

$\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.$

`g,(2,3x -6,9)(0,1x -2)=0`

$\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.$

NL

Nguyễn Linh

3 tháng 2 2022

Giải các phương trình sau:

a/ (3x – 2)(4x + 5) = 0
b/ (2,3x – 6,9)(0,1x + 2) = 0
c/ (4x + 2)(x² + 1) = 0
d/(2x + 7)(x – 5)(5x + 1) = 0
e/ (x – 1)(2x + 7)(x² + 2) = 0
f/ (3x + 2)(x² – 1) = (9x² – 4)(x + 1)

1

NT

Nguyễn Thái Thịnh

3 tháng 2 2022

a) $\left(3x-2\right)\left(4x+5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.$

Vậy: $S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}$

b) $\left(2,3x-6,9\right)\left(0,1x+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.$

c) $\left(4x+2\right)\left(x^2+1\right)=0$

Vì $x^2+1\ge1>0\forall x$

$\Rightarrow4x+2=0$

$\Leftrightarrow x=-\dfrac{1}{2}$

Vậy: $S=\left\{-\dfrac{1}{2}\right\}$

d) $\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.$

Vậy: $S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}$

e) $\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0$

Vì $x^2+2\ge2>0\forall x$

$\Rightarrow\left(x-1\right)\left(2x+7\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.$

f) $\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)$

$\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0$

$\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0$

$\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0$

$\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0$

$\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.$

Vậy: $S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}$

Bài 2: giải phương trình sau a) $X^4$-$x^2$-2=0b) (x+1)$^4$-x$^2$+2)$^2$=0c)3x$^2$-2x-8=0Bài 3: giải phương trình sau a) x$^3$-0,25=0b) x$^4$+2x$^3$+x$^2$=0c) x$^3$-1=0d) 6x$^2$-7x+2=0Mong có người giải giùm xin kẻm ơn...

P

Panda 卐

5 tháng 2 2022

1

NL

Nguyễn Lê Phước Thịnh

5 tháng 2 2022

Bài 3:

b: $\Leftrightarrow x^2\left(x+1\right)^2=0$

hay $x\in\left\{0;-1\right\}$

c: $\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0$

=>x-1=0

hay x=1

d: $\Leftrightarrow6x^2-3x-4x+2=0$

$\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0$

hay $x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}$

Đúng(3)

NN

Ngô Ngọc Tâm Anh

16 tháng 2 2022

3.15 giải các phương trình sau :

a) ( x - 6 ) ( 2x - 5 ) ( 3x + 9 ) = 0

b) 2x( x - 3 ) + 5( x - 3 ) = 0

c) ( x^2 - 4 ) - ( x - 2 ) ( 3 - 2x ) =0

3.16 tìm m để phương trình sau có nghiệm :

x=-7 ( 2m - 5 )x - 2m^2 + 8

3.17 giải các phương trình sau :

a) ( 2x - 1 )^2 - ( 2x + 1 ) = 0

2

MY

missing you =

16 tháng 2 2022

$a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.$

$b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.$

$c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.$

$x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2$

$ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}$$\Rightarrow x=\dfrac{8-2m^2}{14m-34}$

$3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.$

DT

Đỗ Tuệ Lâm

16 tháng 2 2022

3.15:

a, $\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.$

b, $\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.$

c, $\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0$

$\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.$

3.16

$\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0$

$\Leftrightarrow-14m+35-2m^2+8=0$

$\Leftrightarrow-14m-2m^2+43=0$

$\Leftrightarrow-2\left(7m+m^2\right)=-43$

$\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}$

$\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0$

$\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0$

pt vô nghiệm

giải phương trình qui về phương trình...

S

Shreya

20 tháng 3 2023

3

NL

Nguyễn Lê Phước Thịnh

20 tháng 3 2023

a: =>x(x+3)=0

=>x=0 hoặc x=-3

b: =>x(1-2x)=0

=>x=0 hoặc x=1/2

c: =>(x-7)(2x+3-x)=0

=>(x-7)(x+3)=0

=>x=7 hoặc x=-3

d: =>(x-2)(3x-1-x-3)=0

=>(x-2)(2x-4)=0

=>x=2

NH

Ngô Hải Nam

20 tháng 3 2023

a)

`x^2 +3x=0`

`<=>x(x+3)=0`

$< =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.$

b)

`x-2x^2 =0`

`<=>x(1-2x)=0`

$< =>\left[{}\begin{matrix}x=0\\1-2x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.$

c)

`(x-7)(2x+3)=x(x-7)`

`<=>(x-7)(2x+3)-x(x-7)=0`

`<=>(x-7)(2x+3-x)=0`

`<=>(x-7)(x+3)=0`

$< =>\left[{}\begin{matrix}x-7=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.$

d)

`(x-2)(x+3)=(x-2)(3x-1)`

`<=>(x-2)(x+3)-(x-2)(3x-1)=0`

`<=>(x-2)(x+3-3x+1)=0`

`<=>(x-2)(-2x+4)=0`

$< =>\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\\ < =>x=2$

L

Lyna

25 tháng 2 2022

1,Giải các phương trình sau:

a,(x-4)^2-25 = 0

b,(x-3)^2-(x+1)^2 = 0

c,(x^2-4)(2x-3) = (x^2 - 4 )(x-1)

d,(3x-7)^2 - 4( x+1)^2 = 0

Giải giúp em với ạ :3

2

NH

Nguyễn Huy Tú

25 tháng 2 2022

a,$\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1$

b, $\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1$

c, $\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0$

$\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2$

d, $\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0$

$\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9$

BN

Bùi Ngọc Yến

25 tháng 2 2022

a) Ta có: 4x-20=0

$\Leftrightarrow 4 x = 20$

hay x=5

Vậy: S={5}

b) Ta có: $2 x + x + 12 = 0$

$\Leftrightarrow 3 x + 12 = 0$

$\Leftrightarrow 3 x = - 12$

hay x=-4

Giải các phương trình sau:a, x2 - 9x +20 = 0b, x2 - 3x - 18 = 0c, 2x2 - 9 x + 9 = 0d, 3x2 - 8x + 4 = 0e, 3x3 - 6x2 - 9x = 0f, x(x - 5) - 2 + x = 0g, x3 + 32 + 6x +8 = 0h, 2x(x - 2) - 2 + x = 0i, 5x(1 - x) + x - 1 = 0k, 4 - 9(x - 1)2 = 0l, (x - 2)2 - 36(x + 3)2 =...

D

DakiDaki

18 tháng 2 2022

2

RH

Rin Huỳnh

18 tháng 2 2022

$a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}$

NL

Nguyễn Lê Phước Thịnh

18 tháng 2 2022

d: $\Leftrightarrow3x^2-6x-2x+4=0$

=>(x-2)(3x-2)=0

=>x=2 hoặc x=2/3

e: $\Leftrightarrow3x\left(x^2-2x-3\right)=0$

=>x(x-3)(x+1)=0

hay $x\in\left\{0;3;-1\right\}$

f: $\Leftrightarrow x^2-5x-2+x=0$

$\Leftrightarrow x^2-4x-2=0$

$\Leftrightarrow\left(x-2\right)^2=6$

hay $x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}$

DN

dũng nguyễn đăng

5 tháng 9 2021

Giải các phương trình sau:
a) x^2 – 4x + 4 = 25
b) (5 – 2x)^2 – 16 = 0
c) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15

4

I

ILoveMath

5 tháng 9 2021

a) $x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.$

b) $\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.$

c) $\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.$

弃佛入魔

5 tháng 9 2021

a)$\Leftrightarrow$$x^2-4x-21=0$

$\Leftrightarrow$$x^2-7x+3x-21=0$

$\Leftrightarrow$$x(x-7)+3(x-7)=0$

$\Leftrightarrow$$(x-7)(x+3)=0$

$\Leftrightarrow$$\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.$

b)$\Leftrightarrow$$(5-2x)^2-4^2=0$

$\Leftrightarrow$$(5-2x-4)(5-2x+4)=0$

$\Leftrightarrow$$(-2x+1)(-2x+9)=0$

$\Leftrightarrow$$\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.$

Bài1: giải các pt sau:a, 3-4x+24+6x= x+27+3x b, 5-(6-x)=4(3-2x)c, x-(x+1)/3 = (2x+1)/5d,(2x-1)/3 - (5x+2)/7 = x+13Bài 2:a, (x-1)(3x+1)=0b, (x-5)(7-x)=0c, ( x-1)(x+5)(-3x+8)=0d, x(x^2 - 1 )=0Giúp mình 2 bài này với , mình đang cần gấp , CẢM ƠN M.N...

HN

Hoàng Ngân Nguyễn

18 tháng 4 2023

2

NL

Nguyễn Lê Phước Thịnh

19 tháng 4 2023

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>$\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}$

d: =>x=0 hoặc x^2-1=0

=>$x\in\left\{0;1;-1\right\}$

H

⭐Hannie⭐

18 tháng 4 2023

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