Làm ngắn gọn thôi nhé :v

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

\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)

\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)

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

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

\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)

\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)

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

\(B=\frac{x+2}{x-2}\)

\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{10x}{-x^2+9}\)

\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)

\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)

\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)

\(D=\frac{51x-15}{2x^3-18x}\)

\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)

\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)

\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)

\(E=\frac{10x^2+10}{x^4-2x+1}\)

\(a,\frac{-9}{x}=\frac{-9}{\frac{4}{49}}\)

\(\Rightarrow x=\frac{4}{49}\)

\(b,\left|x-2\right|+\left|x+3\right|=0\)

\(\left|x-2\right|\ge0;\left|x+3\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-2\right|=0\\\left|x+3\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-3\end{cases}vl}}\)

\(c,3x^2+9x+6=0\)

\(\Rightarrow3x^2+3x+6x+6=0\)

\(\Rightarrow3x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Rightarrow\left(3x+6\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+6=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)

\(d,x^2-7x-8=0\)

\(\Rightarrow x^2+x-8x-8=0\)

\(\Rightarrow x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Rightarrow\left(x-8\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)

Sửa đề: x² + 13x + 41 --> x² + 13x + 42

Giải:

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)

(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)

\(\Leftrightarrow x^2+8x+7=12\)

⇔$x^2-8x=5$

⇔ $8x+{(-4)}^2=5+{(-4)}^2$
⇔ $x^2-8x+16=21$
⇔ ${(x-4)}^2=21$
⇔ $x=\pm \sqrt{21}+4$

Vậy...

Chúc bạn học tốt@@

vabh ơi cho mk hỏi bạn có ghi sai đề k ạ?

sr mik viết sai đề 42 ko phải 41

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

h.3x - 2/6 - 5 = 3 - 2(x + 7)/4

<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4

<=> 3x - 32/6 = 3 - 2x - 14/4

<=> 3x - 32/6 = -2x - 11/4

<=> 6x - 64/12 = -6x - 33/12

<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12

a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)

Vậy \(x=\frac{-3}{4}\)

b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)

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

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

a)\(M=\left(1-\frac{6-2x^3}{x^6-9}\right).\frac{4}{x^5+3x^2}:\left[\frac{6x^6-24}{x^9+6x^6+9x^3}:\left(\frac{3x^2}{2}+\frac{3}{x}\right)\right]\)

\(=\left(1-\frac{-2\left(x^3-3\right)}{\left(x^3+3\right)\left(x^3-3\right)}\right).\frac{4}{x^2\left(x^3+3\right)}:\left[\frac{6\left(x^3-2\right)\left(x^3+2\right)}{x^3\left(x^3+3\right)^2}:\frac{3x^3+6}{2x}\right]\)

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

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

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