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P
Phong
CTVHS
7 tháng 7 2024

Bạn viết rõ lại đề 

10 tháng 9 2024

\(\dfrac{1}{3x}-\dfrac{1}{4x}=\dfrac{1}{x^2}\)

\(\dfrac{4x-3x}{4x3x}=\dfrac{1}{x^2}\)

\(\dfrac{1}{12x}=\dfrac{1}{x^2}\)

\(\Rightarrow12x=x.x\)

\(\Rightarrow12=x\)

Vậy \(x=12\)

\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{3}{4x^2}\)

\(\dfrac{4x-3x}{3.2}=\dfrac{3}{4x^2}\)

\(\dfrac{1}{2}=\dfrac{1}{4x}\)

\(\Rightarrow2=4x\)

\(\Rightarrow x=\dfrac{2}{4}\)

Vậy \(x=\dfrac{2}{4}\)

 

3 tháng 2 2017

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

14 tháng 12 2018

\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

14 tháng 12 2018

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)

8 tháng 6 2019

Tìm x:

1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)

Vậy x = 5

2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)

\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)

\(\Leftrightarrow-4x+15=-7\)

\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)

Vậy x = \(\frac{11}{2}\)

3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6

\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)

\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)

Vậy x = -1

4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)

\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)

\(\Leftrightarrow14x=0\Leftrightarrow x=0\)

Vậy x = 0

5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)

\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)

6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27

\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)

\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)

\(\Leftrightarrow-x^3=27\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

7. 3x (8x - 4) - 6x (4x - 3) = 30

\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)

\(\Leftrightarrow0=30\) ( vô lý)

Vậy pt vô nghiệm

8. 3x (5 - 2x) + 2x (3x - 5) = 20

\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)

\(\Leftrightarrow5x=20\Leftrightarrow x=4\)

Vậy x = 4

3 tháng 6 2018

1.

a) \(\left\{4x-2\left(x-3\right)-3\left[x-3\left(4-2x\right)+8\right]\right\}.\left(-3x\right)\)

= \(\left[4x-2x+6-3\left(x-12+6x\right)+8\right].\left(-3x\right)\)

\(=\left(4x-2x+6-3x+36-18x+8\right).\left(-3x\right)\)

= \(\left(-19x+50\right).\left(-3x\right)\)

\(=57x^2-150x\)

b) \(5\left(3x^2+4y^3\right)+\left[9\left(2x^2-y^3\right)-2\left(x^2-5y^3\right)\right]\)

\(=15x^2+20y^3+\left(18x^2-9y^3-2x^2+10y^3\right)\)

\(=15x^2+20y^3+16x^2+y^3\)

\(=31x^2+21y^3\)

2.

a) \(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

\(\Rightarrow5x-10x^2-3x^2-54x=0\)

\(\Rightarrow-49x-13x^2=0\)

\(\Rightarrow x\left(-49-13x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-49}{13}\end{matrix}\right.\)

b)

\(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)

\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)

\(\Rightarrow5x-12x+24x-90x+36=182\)

\(\Rightarrow-73x-146=0\)

\(\Rightarrow x=-2\)

3 tháng 6 2018

cảm ơn bạnvui

14 tháng 3 2020

1) \(\frac{4x-8}{2x^2+1}=0\)

<=> \(\frac{4\left(x-2\right)}{2x^2+1}=0\)

<=> 4(x - 2) = 0

<=> x - 2 = 0

<=> x = 2

2) \(\frac{x^2-x-6}{x-3}=0\)

<=> \(\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

<=> x + 2 = 0

<=> x = -2

3) xem ở đây Câu hỏi của Vương Thanh Thanh

4) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

<=> \(\frac{12}{\left(1+3x\right)\left(1-3x\right)}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

<=> 12 = (1 - 3x)2 - (1 + 3x2)

<=> 12 = 1 - 6x + 9x2 - 1 - 6x - 9x2

<=> 12 = -12x

<=> x = -1

5) ĐKXĐ: \(x\ne1,x\ne3\)

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

<=> \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

<=> (x + 5)(x - 3) = (x + 1)(x - 1) - 8

<=> x2 - 3x + 5x - 15 = x2 - x + x - 1 - 8

<=> x2 + 2x - 15 = x- 9

<=> x2 + 2x - 15 - x2 = -9

<=> 2x - 15 = -9

<=> 2x = -9 + 15

<=> 2x = 6

<=> x = 3 (ktm)

=> pt vô nghiệm

6) ĐKXĐ: \(x\ne\pm2\)

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

<=> \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)

<=> (x + 1)(x + 2) - 5(x - 2) = 12 + (x - 2)(x + 2)

<=> x2 + 2x + x + 2 - 5x + 10 = 12 + x2 + 2x - 2x - 4

<=> x2 - 2x + 12 = x2 + 8

<=> x2 - 2x + 12 - x2 = 8

<=> -2x + 12 = 8

<=> -2x = 8 - 12

<=> -2x = -4

<=> x = 2 (ktm)

=> pt vô nghiệm

26 tháng 7 2018

uầy ,bn cứ nhân vào là đc

26 tháng 7 2018

Thôi giúp luôn  =.=

\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)

\(=x^2+6x+9+6x^2-10x+3x-5-6x+2x^2+4x+25\)

\(=9x^2-3x+29\)

P
Phong
CTVHS
27 tháng 6 2023

a) \(\sqrt{1-4x+4x^2}=5\) 

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2=3x-1}\)

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

\(\Leftrightarrow x+3=3x-1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

27 tháng 6 2023

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)

9 tháng 10 2017

Giúp mk với, mai có rồi

13 tháng 10 2017

bn học sách vnen hay sách cũ z

30 tháng 9 2016

Bài 2

a) 4x(x-3)-3x+9

=4x(x-3)-3(x-3)

= (x-3)(4x-3)

b) x3+2x2-2x-4

=(x3+2x2)-(2x+4)

=x2(x+2)-2(x+2)

=(x+2)(x2-2)

c) 4x2-4y+4y-1

=4x2-1

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

d) x5-x

=x(x4-1)

=x(x2-1)(x2+1)

31 tháng 10 2018

a) 4x(x-3)-3x+9

= 4x(x-3) - 3(x-3)

= (x-3)(4x-3)

b)x3 + 2x2 - 2x - 4

= x2(x + 2) - 2(x + 2)

= (x+2)(x2-2)

c) 4x2 - 4y +4y -1

= [(2x)2-12] + (-4y+4y)

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

d) x5-x

= x(x4 - 1)