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22 tháng 1 2021

\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)

\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)

\(\Leftrightarrow4\left(x^2-2x+10x-20\right)-\left(x^2+4x+10x+40\right)=3\left(x^2-2x+4x-8\right)\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)

\(\Leftrightarrow3x^2+18x-120=3x^2+6x-24\)

\(\Leftrightarrow12x=96\)\(\Leftrightarrow x=8\)

Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)

22 tháng 1 2021

\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40-3x^2-6x+24=0\)

\(\Leftrightarrow12x-96=0\)

\(\Leftrightarrow x=8\)

29 tháng 1 2018

A) Ta có: \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)

\(\Leftrightarrow4x^2-x^2-3x^2+32x-14x-6x=-24+80+40\)

\(\Leftrightarrow12x=96\)

\(\Leftrightarrow x=8\)

Vậy x = 8

B) Ta có: \(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)

\(\Leftrightarrow\left(x+2\right)^2-2.8\left(2x+1\right)=25.8+\left(x-2\right)^2\)

\(\Leftrightarrow x^2+4x+4-32x-16=200+x^2-4x+4\)

\(\Leftrightarrow x^2-x^2+4x-32x+4x=200+4-4+16\)

\(\Leftrightarrow-24x=216\)

\(\Leftrightarrow x=-9\)

Vậy x = -9

27 tháng 9 2020

999+2819=

27 tháng 3 2020

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

8 tháng 1 2020

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

2 tháng 4 2017

\(\left(x-1\right)\left(x+1\right)-2\left(2x+3\right)\le\left(x-2\right)^2+x\)

\(\Leftrightarrow x^2-1-4x-6\le x^2-4x+4+x\)

\(\Leftrightarrow x^2-4x-7\le x^2-3x+4\)

\(\Leftrightarrow x^2-4x-x^2+3x\le7+4\)

\(\Leftrightarrow-x\le11\)

\(\Leftrightarrow x\le-11\)

2 tháng 4 2017

biết đừng đăng anh à

11 tháng 3 2020

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5

<=> 7x2 + 2x - 7x2 + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

11 tháng 3 2020

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80

<=> 4x2 + 20x - 4x2 - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

23 tháng 3 2020

AYUASGSHXHFSGDB HAGGAHAJF

27 tháng 5 2020

d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne-2;x\ne-3\)

\(\Leftrightarrow x+3+x+2=1\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\) (không nhận)

Vậy : \(S=\varnothing\)

27 tháng 5 2020

Giai phương trình sau :

a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)

ĐKXĐ : \(x\ne1;x\ne-5\)

Với điều kiện trên ta có :

\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)

\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)

\(\Leftrightarrow10-3x-15=5x-5\)

\(\Leftrightarrow-8x=0\)

\(\Leftrightarrow x=0\) (nhận)

Vậy : \(S=\left\{0\right\}\)

8 tháng 8 2019

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!