\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)

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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)

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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)

\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)\)

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

\(\Leftrightarrow-x^2+12x+4=16x+4\)

\(\Leftrightarrow x^2+4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-4\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

qui đồng là xong lười làm hả

a) \(\Leftrightarrow\dfrac{15x}{x^2+3x-4}-1=\dfrac{12}{x+4}+\dfrac{4}{x-1}\)

\(\Leftrightarrow\dfrac{15x}{x^2+4x-x-4}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)

\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-\dfrac{12}{x+4}-\dfrac{4}{x-1}=1\)

\(\Leftrightarrow\dfrac{15x-12x+12-4x-16}{\left(x-1\right)\left(x+4\right)}=1\)

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

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

\(\Rightarrow x=0\)

tick cho t vs hik

b) \(\Leftrightarrow\left|x-2\right|+3=5\)

\(\Leftrightarrow\left|x-2\right|=5-3\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

\(\Rightarrow\dfrac{15x}{x^2+3x-4}-1=\dfrac{12}{x+4}+\dfrac{12}{3x-3}\)

\(\Rightarrow\dfrac{15}{x^2+3x-4}-1=\dfrac{36x-36}{\left(x+4\right)\left(3x-3\right)}+\dfrac{12x+48}{\left(x+4\right)\left(3x-3\right)}\)

\(\Rightarrow\dfrac{15}{x^2+3x-4}-1=\dfrac{36x-36+12x+48}{\left(x+4\right)\left(3x-3\right)}\)

\(\Rightarrow\dfrac{15}{x^2+3x-4}-1=\dfrac{48x+12}{x\left(3x-3\right)+4\left(3x-3\right)}\)

\(\Rightarrow\dfrac{15}{x^2+3x-4}-1=\dfrac{48x+12}{3x^2-3x+12x-12}\)

\(\Rightarrow\dfrac{15}{x^2+3x-4}=\dfrac{48x+12}{3x^2+9x-12}\)

\(\Rightarrow15\left(3x^2+9x-12\right)=\left(48x+12\right)\left(x^2+3x-4\right)\)

Nhân ra rồi tính là ok

b)

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

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

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

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

\(\Leftrightarrow2x^2-8=-4x^2+4\) ( điều kiện \(x\ne\pm1,x\ne\pm2\) )

\(\Leftrightarrow6x^2=12\)

\(\Rightarrow x=\pm\sqrt{2}\)

a )

\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

\(\Leftrightarrow\dfrac{15x-\left(x^2+3x-4\right)}{x^2+3x-4}=\dfrac{12}{x+4}+\dfrac{12}{3x-3}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{x^2+4x-x-4}=\dfrac{48x+12}{\left(x+4\right)\left(3x-3\right)}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{x\left(x+4\right)-\left(x+4\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{\left(x+4\right)\left(x-1\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)

\(\Leftrightarrow12x-x^2+4=\dfrac{48x+12}{3}\)

\(\Leftrightarrow12x-x^2+4=16x+4\)

\(\Leftrightarrow x^2+8x=0\)

\(\Delta=b^2-4ac\)

\(\Delta=64\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-8+\sqrt{64}}{2}=0\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-8-\sqrt{64}}{2}=-8\left(loại\right)\end{matrix}\right.\)

Do \(x=-8\) không thỏa mãn phương trình

Vậy \(x=0\)

c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

a: =>2(2x-3)-9=5-3x-2

=>4x-6-9=-3x+3

=>4x-15=-3x+3

=>7x=18

=>x=18/7

b: =>\(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\dfrac{21}{3x}+2\)

=>\(\dfrac{23}{3x}=\dfrac{4}{5}+2+\dfrac{1}{4}=\dfrac{61}{20}\)

=>3x=460/61

=>x=460/183