K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

27 tháng 2 2019

a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)

\(\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1\)

\(\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}\)

\(\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0\)

\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow2015-x=0\)

<=> x=2015

Vậy phương trình có nghiệm là x=2015

27 tháng 2 2019

b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)

\(\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1\)

\(\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}\)

\(\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0\)

\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x-2015=0\)

=> x=2015

Vậy phương trình có nghiệm x=2015

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

AH
Akai Haruma
Giáo viên
5 tháng 3 2021

Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!

NV
18 tháng 3 2021

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

NV
18 tháng 3 2021

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

30 tháng 4 2018

a) 4x -8 ≥ 3(3x-1)-2x +1

⇒4x -8 ≥7x -2

⇒4x -7x ≥ -2 +8

⇒-3x ≥ 6

⇒x≤-2

Vậy bpt có nghiệm là:{x|x≤-2}

30 tháng 4 2018

b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4

⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4

⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16

⇔ -3x ≤ -6

⇔ x≥ 2

Vậy bpt có tập nghiệm là: {x|x≥2}

a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)

\(\Leftrightarrow4x-2+2x=5x-20\)

\(\Leftrightarrow x=-18\)

b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

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

hay \(x=\dfrac{1}{3}\)

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

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

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

29 tháng 8 2021

undefined

1 tháng 2 2019

a)MTC 15

\(\dfrac{\left(x-3\right)\times3}{15}=\dfrac{6.15-\left(1-2x\right)\times5}{15}=\dfrac{3x-9}{15}=\dfrac{90-5-10x}{15}=3x-9=90-5-10x\Leftrightarrow3x+10x=90-5+9\)

1 tháng 2 2019

Chưa nghỉ tết à :))

\(a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Rightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-10x=90-5+9\)

\(\Leftrightarrow-7x=94\)

\(\Leftrightarrow x=-\dfrac{94}{7}\)

Vậy.....

\(b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Rightarrow2\left(3x-2\right)-5.12=3\left[3-2\left(x+7\right)\right]\)

\(\Leftrightarrow6x-4-60=-6x-33\)

\(\Leftrightarrow6x+6x=-33+60+4\)

\(\Leftrightarrow12x=31\)

\(\Leftrightarrow x=\dfrac{31}{12}\)

Vậy.....

\(c,2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)

\(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

\(\Leftrightarrow2x+x=5-\dfrac{13}{5}-\dfrac{6}{5}\)

\(\Leftrightarrow3x=\dfrac{6}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy.....

\(d,\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)

\(\Rightarrow28\left[5\left(x-1\right)+2\right]-42\left(7x-1\right)=24\left[2\left(2x+1\right)\right]-5.168\)

\(\Leftrightarrow140x-84-294x+42=96x+48-840\)

\(\Leftrightarrow140x-294x-96x=48-840-42+84\)

\(\Leftrightarrow-250x=-750\)

\(\Leftrightarrow x=3\)

Vậy.....

\(e,\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)

\(\Rightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left[2\left(x-1\right)\right]\)

\(\Leftrightarrow6x-6+3x-3=12-8x+8\)

\(\Leftrightarrow6x+3x+8x=12+8+3+6\)

\(\Leftrightarrow17x=29\)

\(\Leftrightarrow x=\dfrac{29}{17}\)

Vậy.....

\(g,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow\dfrac{2}{2001}-\dfrac{x}{2001}-1=\dfrac{1}{2002}-\dfrac{x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow-\dfrac{x}{2001}+\dfrac{x}{2002}+\dfrac{x}{2003}=\dfrac{1}{2002}+1-\dfrac{2}{2001}\)

\(\Leftrightarrow x\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)=1+\dfrac{1}{2002}-\dfrac{2}{2001}\)

\(\Leftrightarrow x=\dfrac{\left(1+\dfrac{1}{2002}-\dfrac{2}{2001}\right)}{\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)}=2003\)

Vậy.....

ai bít thì giúp mình với nhé

\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)

\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)

\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)

\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)

\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)

\(\Leftrightarrow2015-x=0\)

\(\Leftrightarrow x=2015\)

KL : PT có nghiệm \(S=\left\{2015\right\}\)

NV
5 tháng 4 2021

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:

\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)

\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)

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

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

NV
5 tháng 4 2021

Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý

b.

Đặt \(x+3=t\) 

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)

\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)

\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)

Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.

22 tháng 5 2017

a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)

\(\Leftrightarrow\dfrac{4x+\left(2x-1\right)}{6}=\dfrac{24-2x}{6}\)

\(\Leftrightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow6x+2x=24+1\)

\(\Leftrightarrow8x=25\)

\(\Leftrightarrow x=\dfrac{25}{8}\)

Vậy phương trình có một nghiệm là x = \(\dfrac{25}{8}\)

b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)

\(\Leftrightarrow\dfrac{6\left(x-1\right)+3\left(x-1\right)}{12}=\dfrac{12-8\left(x-1\right)}{12}\)

\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)

\(\Leftrightarrow9\left(x-1\right)+8\left(x-1\right)=12\)

\(\Leftrightarrow17\left(x-1\right)=12\)

\(\Leftrightarrow17x-17=12\)

\(17x=12+17\)

\(\Leftrightarrow17x=29\)

\(\Leftrightarrow x=\dfrac{29}{17}\)

Vậy phương trình có một nghiệm là x = \(\dfrac{29}{17}\)

c) \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow\dfrac{2-x}{2001}-\dfrac{1-x}{2002}-\dfrac{\left(-x\right)}{2003}=1\)

\(\Leftrightarrow\dfrac{2-x}{2001}+1-\dfrac{1-x}{2002}-1-\dfrac{\left(-x\right)}{2003}-1=1+1-1-1\)

\(\Leftrightarrow\dfrac{2-x}{2001}+\dfrac{2001}{2001}-\dfrac{1-x}{2002}-\dfrac{2002}{2002}-\dfrac{\left(-x\right)}{2003}-\dfrac{2003}{2003}=0\)

\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)

\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow2003-x=0\)

\(\Leftrightarrow-x=-2003\)

\(\Leftrightarrow x=2003\)

Vậy phương trình có một nghiệm là x = 2003

29 tháng 5 2017

a) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\)

\(\Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}=\dfrac{24}{6}-\dfrac{2x}{6}\)

\(\Leftrightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow4x+2x+2x=1+24\)

\(\Leftrightarrow8x=25\)

\(\Leftrightarrow x=\dfrac{25}{8}\)

Vậy S={\(\dfrac{25}{8}\)}

b) \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)

\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}+\dfrac{3\left(x-1\right)}{12}=\dfrac{12}{12}-\dfrac{8\left(x-1\right)}{12}\)

\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-8\left(x-1\right)\)

\(\Leftrightarrow6x-6+3x-3=12-8x+8\)

\(\Leftrightarrow6x+3x+8x=6+3+12+8\)

\(\Leftrightarrow17x=29\)

\(\Leftrightarrow x=\dfrac{29}{17}\)

Vậy S={\(\dfrac{29}{17}\)}

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19