Tim so nguyen x biet:a,x-(-24)=3b,(x 5)-(x-9)=x 3c,c(x 2)=0d,(x-1)^2=0e,x^2-3x=0f,-13-(6-|x 1|)=24g,26-|x 9|=-13i,(3x-6).3=3^4

2

NH

Nguyễn Hoàng Minh

31 tháng 10 2021

$a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.$

T

tholauyeu

31 tháng 10 2021

a.

4x(x-5) - (x-1)(4x-3)-5=0

4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

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

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

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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DD

Di Di

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`

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

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

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

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KH

khánh huyền

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0
b) 3x (x-1) + x - 1 = 0
c) 2(x+3) - x ² - 3x = 0
d) x(x - 2) + 3x - 6 = 0
e) 4x ² - 4x +1 = 0
f) x +5x ² = 0
g) x ² 2x -3 = 0

4

PS

Phía sau một cô gái

30 tháng 7 2021

a) $\left(2x-1\right)^2-25=0$

⇔ $\left(2x-1\right)^2-5^2=0$

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

⇒ $2x-1-5=0$ hoặc $2x-1+5=0$

⇔ $x=3$ hoặc $x=-2$

TC

Thùy Cái

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² = 25

<=> 2x-1 = 5 hay 2x-1 =-5

<=> 2x= 6 hay 2x=-4

<=> x=3 hay x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

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

<=> x-1=0 hay 3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=$\dfrac{-1}{3}$

Vậy S={1;$\dfrac{-1}{3}$}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3 hay x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=$\dfrac{1}{2}$

Vậy S={$\dfrac{1}{2}$}
f) x +5x² = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= $\dfrac{-1}{5}$

Vậy S={0;$\dfrac{-1}{5}$}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

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

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

<=> x-1=0 hay x+3=0

<=> x=1 hay x=-3

Vậy S={1;-3}

VN

vân nguyễn

28 tháng 7 2021

tìm x biết:

a) x - 3= (3 - x)^2

b) x^3 + 3/2x^2 + 3/4x + 1/8 = 1/64

c) 8x^3 - 50x = 0

d) (x - 2) (x^2 + 2x + 7) + 2(x^2 - 4) - 5(x - 2) = 0

e) x(x + 3) - x^2 - 3x = 0

f) x^3 + 27 + (x + 3) (x - 9) = 0

Giúp mik vs ạ

2

NL

Nguyễn Lê Phước Thịnh

29 tháng 7 2021

a) Ta có: $\left(x-3\right)=\left(3-x\right)^2$

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

$\Leftrightarrow\left(x-3\right)\left(x-4\right)=0$

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

b) Ta có: $x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}$

$\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}$

$\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3$

$\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}$

hay $x=-\dfrac{1}{4}$

c) Ta có: $8x^3-50x=0$

$\Leftrightarrow2x\left(4x^2-25\right)=0$

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

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

NL

Nguyễn Lê Phước Thịnh

29 tháng 7 2021

e) Ta có: $x\left(x+3\right)-x^2-3x=0$

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

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

f) Ta có: $x^3+27+\left(x+3\right)\left(x-9\right)=0$

$\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0$

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

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

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

HV

Hưng Việt Nguyễn

5 tháng 9 2021

c)3x^2-7x-10=0

d)2x(x-10)-x+10=0

e)3x^3+7x^2+17x+5=0

f)(2x-1)^2-(x-3)^2=0

g)x^3-5x^2+8x=4

8

HV

Hưng Việt Nguyễn

5 tháng 9 2021

tìm x nha

bài 19: tìm xc) ( 34 - 2x ) . ( 2x - 6 ) = 0d) ( 2019 - x ) . ( 3x - 12 ) 0e) 57 . ( 9x - 27 ) = 0f) 25 + ( 15 - x ) = 30g) 43 - ( 24 - x ) = 20h) 2 . ( x - 5 ) - 17 = 25i) 3 . ( x + 7 ) - 15 = 27j) 15 + 4 . ( x - 2 ) = 95k) 20 - ( x + 14 ) = 5l) 14 + 3 . ( 5 - x ) = 27nhanh nha, mik tick cho, ccau trình bày dễ hiểu, ko cần...

HP

Hồng Phúc

5 tháng 9 2021

c, $3x^2-7x+10=0$

$\Leftrightarrow3x^2+3x-10x+10=0$

$\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.$

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PP

Phượng Phạm

9 tháng 7 2023

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#Toán lớp 6

1

H9

HT.Phong (9A5)

9 tháng 7 2023

c) $\left(34-2x\right)\left(2x-6\right)=0$

$\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.$

d) $\left(2019-x\right)\left(3x-12\right)=0$

$\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.$

e) $57\left(9x-27\right)=0$

$\Rightarrow9x-27=0$

$\Rightarrow9\left(x-3\right)=0$

$\Rightarrow x-3=0$

$\Rightarrow x=3$

f) $25+\left(15-x\right)=30$

$\Rightarrow25+15-x=30$

$\Rightarrow40-x=30$

$\Rightarrow x=40-30$

$\Rightarrow x=10$

g) $43-\left(24-x\right)=20$

$\Rightarrow43-24+x=20$

$\Rightarrow19+x=20$

$\Rightarrow x=20-19$

$\Rightarrow x=1$

h) $2\left(x-5\right)-17=25$

$\Rightarrow2\left(x-5\right)=17+25$

$\Rightarrow x-5=21$

$\Rightarrow x=21+5$

$\Rightarrow x=26$

i) $3\left(x+7\right)-15=27$

$\Rightarrow3\left(x+7\right)=27+15$

$\Rightarrow x+7=14$

$\Rightarrow x=14-7$

$\Rightarrow x=7$

j) $15+4\left(x-2\right)=95$

$\Rightarrow4\left(x-2\right)=95-15$

$\Rightarrow4\left(x-2\right)=80$

$\Rightarrow x-2=20$

$\Rightarrow x=20+2$

$\Rightarrow x=22$

k) $20-\left(x+14\right)=5$

$\Rightarrow x+14=20-5$

$\Rightarrow x+14=15$

$\Rightarrow x=15-14$

$\Rightarrow x=1$

l) $14+3\left(5-x\right)=27$

$\Rightarrow3\left(5-x\right)=27-14$

$\Rightarrow3\left(5-x\right)=13$

$\Rightarrow5-x=\dfrac{13}{3}$

$\Rightarrow x=5-\dfrac{13}{3}$

$\Rightarrow x=\dfrac{2}{3}$

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NP

Nguyễn Phan Anh

4 tháng 8 2021

Bài 3: Giải các phương trình sau:a, 2x3 - 50x = 0b, 2x (3x - 5) - (5 - 3x)c, 9(3x - 2) = x(2 - 3x)d, (2x - 1)2 - 25 = 0e, 25x2 - 2 = 0f, x2 - 25 = 6x - 9g, 5x(x - 3) - 2x + 6 = 0h, 3x(x - 7) - 2(x - 7) = 0i, 7x2 - 28 = 0j, (2x + 1) + x(2x + 1) = 0k, (x + 2)2 - (x - 2)(x + 2) = 0l, x3 + 5x2 - 4x - 20 = 0m, x2 - 25 + 2(x + 5) = 0n, x3 - 3x + 2 = 0o, x2 - 6x + 8 = 0 p, x2 - 5x - 14 = 0q, (x - 2)2 - (x - 3)(x + 3) = 6r, (2x - 1)2 - (2x + 5)(2x - 5) =...

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0

HN

Huy Nguyen

30 tháng 4 2021

B5:Giải pt:

a)2x$^2$-8=0

b)3x$^3$-5x=0

c)x$^4$+3x$^2$-4=0

d)3x$^2$+6x-9=0

e)$\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}$

g)5x$^4$+6x$^2$-11=0

#Toán lớp 9

2

LL

Linh Linh

30 tháng 4 2021

a. 2x$^2$-8=0

2x$^2$=8

x$^2$=4

x=2

b.3x$^3$-5x=0

x(3x$^2$-5)=0

$\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.$⇔$\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.$

LL

Linh Linh

1 tháng 5 2021

c.x $^{4}$ +3x $^{2}$ -4=0$^{\left(\cdot\right)}$

đặt t=x$^2$ (t>0)

ta có pt: t$^2$+3t-4=0 $^{\left(1\right)}$

thấy có a+b+c=1+3+(-4)=0 nên pt$^{\left(1\right)}$ có 2 nghiệm

t$_1$=1; t$_2$=$\dfrac{c}{a}$=-4

khi t$_1$=1 thì x$^2$=1 ⇒x=$^+_-$1

khi t$_2$=-4 thì x$^2$=-4 ⇒ x=$^+_-$2

vậy pt đã cho có 4 nghiệm x=$^+_-$1; x=$^+_-$2

d)3x$^2$+6x-9=0

thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm

x$_1$=1; x$_2$=$\dfrac{c}{a}=\dfrac{-9}{3}=-3$

e. $\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}$ (ĐK: x#5; x#2 )

⇔$\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}$=$\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}$

⇒2x - x$^2$ + 4 - 2x + 6x - 6x$^2$ + 12 - 6x - 6x +30 = 0

⇔-7x$^2$ - 6x + 46=0

Δ'=b'$^2$-ac = (-3)$^2$ - (-7)$\times$46= 9+53 = 62>0

$\sqrt{\Delta'}=\sqrt{62}$

vậy pt có 2 nghiệm phân biệt

x$_1$=$\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}$

x$_2$=$\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}$

vậy pt đã cho có 2 nghiệm x$_1$=.....;x$_2$=......

câu g làm tương tự câu c

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

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