a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ĐKXĐ : \(x\ne-2;x\ne5\)

\(\frac{7}{x+2}=\frac{3}{x-5}\)

<=> 3(x + 2) = 7(x - 5)

<=> 3x + 6 = 7x - 35

<=> 4x = 41

<=>x = 41/4 (tm)

Vậy x = 41/4 là ngiệm phương trình

b) ĐKXĐ \(x\ne\pm3\)

\(\frac{2x-1}{x+3}=\frac{2x}{x-3}\)

<=> \(\frac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

<=> (2x - 1)(x - 3) = 2x(x + 3)

<=> 2x² - 7x + 3 = 2x² + 6x

<=> 13x = 3

<=> x = 3/13 (tm)

Vậy x = 3/13 là nghiệm phương trình

c) ĐKXĐ : \(x\ne-7;x\ne1,5\)

Khi đó \(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\)

<=> \(\frac{\left(3x-2\right)\left(2x-3\right)}{\left(x+7\right)\left(2x-3\right)}=\frac{\left(6x+1\right)\left(x+7\right)}{\left(x+7\right)\left(2x-3\right)}\)

<=> (3x - 2)(2x - 3) = (6x + 1)(x + 7)

<=> 6x² - 13x + 6 = 6x² + 43x + 7

<=> 56x = -1

<=> x = -1/56 (tm) 

Vậy x = -1/56 là nghiệm phương trình

d) ĐKXĐ : \(x\ne\pm1\)

Khi đó \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

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

<=> (2x + 1)(x + 1) = 5(x - 1)²

<=> 2x² + 3x + 1 = 5x² - 10x + 5

<=> 3x² - 13x + 4 = 0

<=> 3x² - 12x - x + 4 = 0

<=> 3x(x - 4) - (x - 4) = 0

<=> (3x - 1)(x - 4) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

Vậy x \(\in\left\{\frac{1}{3};4\right\}\)là nghiệm phương trình

e) ĐKXĐ : \(x\ne1\)

Khi đó \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)

<=> \(\frac{3x-5}{x-1}=2\)

<=> 3x - 5 = 2(x - 1) 

<=> 3x - 5 = 2x - 2

<=> x = 3 (tm) 

Vậy x = 3 là nghiệm phương trình

f) ĐKXĐ : \(x\ne-1\)

 \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)

<=> \(\frac{3x+2}{x+1}=3\)

<=> 3x + 2 = 3(x + 1)

<=> 3x + 2 = 3x + 3

<=> 0x = 1

<=> \(x\in\varnothing\)

Vậy tập nghiệm phương trình S = \(\varnothing\)

g) ĐKXĐ : \(x\ne2\)

Khi đó \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)

<=>\(\frac{x-2}{x-2}=3\)

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

<=> x - 2 = 3x - 6

<=> -2x = -4

<=> x = 2 (loại) 

Vậy tập nghiệm phương trình S = \(\varnothing\)

h) ĐKXĐ : \(x\ne7\)

Khi đó \(\frac{1}{7-x}=\frac{x-8}{x-7}-8\)

<=> \(\frac{x-7}{x-7}=8\)

<=> x - 7 = 8(x - 7)

<=> x - 7 = 8x - 56

<=> 7x = 49

<=> x = 7 (loại)

Vậy tập nghiệm phương trình S = \(\varnothing\)

i) ĐKXĐ : \(x\ne0;x\ne6\)

Ta có : \(\frac{x+6}{x}=\frac{1}{2}+\frac{15}{2\left(x-6\right)}\)

<=> \(\frac{x+6}{x}-\frac{15}{2\left(x-6\right)}=\frac{1}{2}\)

<=> \(\frac{2\left(x+6\right)\left(x-6\right)}{2x\left(x-6\right)}-\frac{15x}{2x\left(x-6\right)}=\frac{1}{2}\)

<=> \(\frac{2x^2-72-15x}{2x\left(x-6\right)}=\frac{1}{2}\)

<=> 4x² - 144 - 30x = 2x(x - 6) 

<=> 2x² - 18x - 144 = 0

<=> x² - 9x - 72 = 0

<=> x² - 9x + 81/4 - 72- 81/4 = 0

<=> \(\left(x-\frac{9}{2}\right)^2-\frac{369}{4}=0\)

<=> \(\left(x-\frac{9}{2}+\sqrt{\frac{369}{4}}\right)\left(x-\frac{9}{2}-\sqrt{\frac{369}{4}}\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{9}{2}-\sqrt{\frac{369}{4}}\\x=\frac{9}{2}+\sqrt{\frac{369}{4}}\end{cases}}\)(tm)

Vậy x \(\in\left\{\frac{9}{2}-\sqrt{\frac{369}{4}};\frac{9}{2}+\sqrt{\frac{369}{4}}\right\}\)

1,\(x-8=3-2\left(x+4\right)\)

\(\Leftrightarrow\)\(x-8=3-2x-8\)

\(\Leftrightarrow x-8-3+2x+8=0\)

\(\Leftrightarrow3x-3=0\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

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

\(\Leftrightarrow2\left(x+3\right)-3\left(x-1\right)-2=0\)

\(\Leftrightarrow2x+6-3x+3-2=0\)

\(\Leftrightarrow-x+7=0\) \(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy \(S=\left\{1\right\}\)

4.

\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)

\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)

\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)

\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)

a: \(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

=>5x=22

hay x=22/5

b: \(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

hay \(x\in\left\{3;-\dfrac{11}{10}\right\}\)

c: \(\Leftrightarrow x^3+2x^2-5x-10+5x=2x^2+17\)

\(\Leftrightarrow x^3+2x^2-10-2x^2-17=0\)

=>x³=27

=>x=3

d: \(\Leftrightarrow x^3+1-x^3+3x=15\)

=>3x=14

hay x=14/3

a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)

\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)

\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)

\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)

\(\Rightarrow12x=36\)

\(\Rightarrow x=\dfrac{36}{12}\)

Vậy x = 3

b) \(x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)

... Phần còn lại cũng tương tự như vậy thôi

a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)

= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)

b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)

= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

c.Tương tự hai câu trên nka!!

d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))

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

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

= \(\dfrac{x-1}{x+1}\)

Lê Cẩm TúThiên ThảoPhạm Thái DươnMai LinhgGuyoSky SơnTùngKhôi Bùi Mysterious PersonPhong ThầnPhùng Khánh Linhtran nguyen baNguyễn Xuân Sángo quanDƯƠNG PHAN KHDũng NguyễnÁNH DƯƠNGlê thị hương giang

1. \(\frac{3x-7}{5}=\frac{2x-1}{3}\)

<=> 3(3x-7)=5(2x-1)

<=> 9x-21=10x-5

<=> -21+5=10x-9x

<=> x=-16

2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)

<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)

=> 9x-21+4x-2=-96

<=> 13x-23=-96

<=> 13x=-73

<=> x=\(\frac{-73}{13}\)

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

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

=> 15x-5x-5=6x+3

<=> 15x-5x-6x=3+5

<=> 4x=8

<=> x=2

4. \(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5-\left(5-2x\right)}{6}\)

<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)

=> 7-3x+9=24x-48+4x

<=> -3x-24x-4x=-48-7

<=> -31x=-55

<=> x= \(\frac{55}{31}\)

5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)

=> 14x-7-15x-6=21x+273

<=> 14x-15x-21x=273+7+6

<=> -22x=286

<=> x= -13

a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)

b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)

c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)