bài 7 tìm x1,x(x 3)-5(x 3)=0 2,5x(x-1)=x-13,(x 1)=(x 1)$^2$ 4,x(2x-3)-2(3-2x)=05,$\left(x-2\right)^2-4=0$ 6,$36x^2=49$7,\(2x\le...

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 12 2022

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

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

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

BÀI 6 tìm x1,$2x\left(x-5\right)-\left(3x+2x^2\right)=0$ 2,$x\left(5-2x\right)+2x\left(x-1\right)=13$3,$2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0$ 4,$5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6$5,$6x^2-\left(2x-3\right)\left(3x+2\right)=1$ ...

TL

thùy linh

16 tháng 12 2022

2

NL

Nguyễn Lê Phước Thịnh

16 tháng 12 2022

1: $\Leftrightarrow2x^2-10x-3x-2x^2=0$

=>-13x=0

=>x=0

2: $\Leftrightarrow5x-2x^2+2x^2-2x=13$

=>3x=13

=>x=13/3

3: $\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0$

=>-2x^2=0

=>x=0

4: $\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6$

=>-8x=6-14=-8

=>x=1

2

2611

16 tháng 12 2022

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

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

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

HA

Hoài An

29 tháng 4 2021

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

2) $5\left(2x-1\right)+7=4\left(2-x\right)+2$

3) $x^2-6x+4\left(x-6\right)=0$

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

4

NL

Nguyễn Lê Phước Thịnh

29 tháng 4 2021

1) Ta có: $\left(3-x^2\right)+6-2x=0$

$\Leftrightarrow3-x^2+6-2x=0$

$\Leftrightarrow-x^2-2x+9=0$

$\Leftrightarrow x^2+2x-9=0$

$\Leftrightarrow x^2+2x+1=10$

$\Leftrightarrow\left(x+1\right)^2=10$

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

Vậy: $S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}$

NL

Nguyễn Lê Phước Thịnh

29 tháng 4 2021

2) Ta có: $5\left(2x-1\right)+7=4\left(2-x\right)+2$

$\Leftrightarrow10x-5+7=8-4x+2$

$\Leftrightarrow10x+4x=8+2+5-7$

$\Leftrightarrow14x=8$

$\Leftrightarrow x=\dfrac{4}{7}$

Vậy: $S=\left\{\dfrac{4}{7}\right\}$

LT

LÊ TRẦN BÁCH

VIP

20 tháng 9 2023

2. tìm x

a) $\left(x-1\right)^3=8$

b) $7^{2x-6}=49$

c) $\left(2x-14\right)^7=128$

d) $x^4.x^5=5^3.5^6$

e) $\left[3.\left(x+2\right):7\right].4=120$

#Toán lớp 6

2

ED

🐟 ⋆ 🐇 🎀 𝒩𝑔𝓊𝓎ễ𝓃 Đă𝓃𝑔 𝒩𝒽â𝓃 🎀 🐇 ⋆ 🐟

20 tháng 9 2023

a) $\left(x-1\right)^3=8=2^3$

$x-1=2$

$x=2+1=3$

b) $7^{2x-6}=49=7^2$

$2x-6=2$

$2x=6+2=8$

$x=8:2=4$

c) $\left(2x-14\right)^7=128=2^7$

$2x-14=2$

$2x=14+2=16$

$x=16:2=8$

d) $x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5$

$x=5$

e) $3\cdot\left(x+2\right):7\cdot4=120$

$x+2=120:3\cdot7:4$

$x+2=70$

$x=70-2=68$

AH

Akai Haruma

Giáo viên

20 tháng 9 2023

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e.

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

1) giải phương trình : a) $\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)$ b) $\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0$ c) $\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}$ d) $\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0$ i) $\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}$ k)...

NT

nguyễn thái hồng duyên

10 tháng 7 2018

1

NL

Nguyễn Lê Phước Thịnh

20 tháng 7 2022

b: $\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0$

$\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0$

$\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0$

=>(7x+10)(x-3)=0

hay $x\in\left\{-\dfrac{10}{7};3\right\}$

d: $\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0$

$\Leftrightarrow26x+91+x^2-9-12x-14=0$

$\Leftrightarrow x^2+14x+68=0$

hay $x\in\varnothing$

1) $\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}$2) $\dfrac{1}{x^2-4}+\dfrac{2}{x+2}< \dfrac{-3}{x-2}$3) (4-x-$3x^2$).(x+2).(x+1) > 04) ($x^3$-9x).(x-3) ≥ 05) $\left|4-x\right|$ ≥ 2x-16) $\left|x-2\right|$ ≤...

MD

Miner Đức

28 tháng 1 2021

#Toán lớp 10

3

HP

Hồng Phúc

30 tháng 1 2021

1.

ĐK: $x\ne3;x\ne-2$

$\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}$

$\Leftrightarrow\dfrac{5\left(x+2\right)+3\left(x-3\right)}{x^2-x-6}\le\dfrac{3+2x}{x^2-x-6}$

$\Leftrightarrow\dfrac{8x+1-3-2x}{x^2-x-6}\le0$

$\Leftrightarrow\dfrac{6x-2}{x^2-x-6}\le0$

$\Leftrightarrow\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.$ hoặc $\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.$

TH1: $\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\-2< x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 3$

TH2: $\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -2$

Vậy ...

HP

Hồng Phúc

30 tháng 1 2021

2.

ĐK: $x\ne\pm2$

$\dfrac{1}{x^2-4}+\dfrac{2}{x+2}>-\dfrac{3}{x-2}$

$\Leftrightarrow\dfrac{1}{x^2-4}+\dfrac{2\left(x-2\right)+3\left(x+2\right)}{x^2-4}>0$

$\Leftrightarrow\dfrac{5x+3}{x^2-4}>0$

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

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

Vậy ...

T

tl:)

18 tháng 3 2022

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

$2,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13$

$3,\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}$

3

MH

Minh Hiếu

18 tháng 3 2022

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

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

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

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

$2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13$

$\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}$

$\Leftrightarrow14x-7-15x-6=21x+273$

$\Leftrightarrow-x-13=21x+273$

$\Leftrightarrow-22x=286$

$\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}$

$3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)$

$\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}$

$\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}$

$\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6$

$\Leftrightarrow x-2=7x-6$

$\Leftrightarrow-6x=-4$

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

KK

Kaito Kid

18 tháng 3 2022

1.(x−5)²+3(x−5)=0

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

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

=>x-5=0 hoặc x-2=0

=>x=5 hoặc x=2

KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

29 tháng 1 2022

Tìm x...

#Toán lớp 6

2

KS

Kudo Shinichi AKIRA^_^

29 tháng 1 2022

Chia nhỏ ra

NL

Nguyễn Lê Phước Thịnh

29 tháng 1 2022

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

bài 1: giải các bất phương trình sau: 1) (x-3)(4-x)≥0 2) $\frac{1+2x}{3x-4}< 0$ 3) (x+1)(x-1)(3x-6)>0 4) 3x(2x+7)(9-3x)≥0 5) $\frac{\left(2x-5\right)\left(x+2\right)}{-4x+3}>0$ 6) $\frac{2}{x-1}\le\frac{5}{2x-1}$ 7) $\frac{x-3}{x+1}\frac{x+5}{x-2}$ 8)...

MM

Mộc Miên

17 tháng 3 2020

#Toán lớp 10

0

H

hmone

6 tháng 10 2021

Bài 1: Tìm x :
a. ( x + 4 )² - ( x + 1 ) ( x - 1 ) = 16
b. ( 2x - 1 )² + ( x + 3 )² - 5 ( x + 7 ) ( x - 7 ) = 0
c. ( x - 2 )³ - ( x - 3 ) ( x² + 3x + 9 ) + 6 ( x + 1 )² = 49