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1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)

ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

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

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

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

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)

<=> \(\frac{3x+10}{x^2+2x-3}=0\)

<=> \(3x+10=0\)

<=> \(x=-\frac{10}{3}\)

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)

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

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

\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow2x+4=2x+3\)

\(\Leftrightarrow0x=-1\)(vô nghiệm)

Vậy phương trình vô nghiệm.

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow2x+7=-10\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)

Trả lời:

a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)\(\left(đkxđ:x\ne-1\right)\)

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

\(\Rightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow4+2x=2x+3\)

\(\Leftrightarrow2x-2x=3-4\)

\(\Leftrightarrow0x=-1\)(không thỏa mãn)

Vậy \(S=\varnothing\)

b, \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\)\(\left(đkxđ:x\ne\frac{3}{2}\right)\)

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

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow x^2+2x+7=x^2-10\)

\(\Leftrightarrow x^2+2x-x^2=-10-7\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(tm)

Vậy \(S=\left\{\frac{-17}{2}\right\}\)

c, \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)\(\left(đkxđ:x\ne1\right)\)

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

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

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

\(\Rightarrow2-5x+2x^2-3x+1=2x-2+2x^2+2x-6\)

\(\Leftrightarrow2x^2-8x+3=2x^2+4x-8\)

\(\Leftrightarrow2x^2-8x-2x^2-4x=-8-3\)

\(\Leftrightarrow-12x=-13\)

\(\Leftrightarrow x=\frac{13}{12}\)(tm)

Vậy \(S=\left\{\frac{13}{12}\right\}\)

Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.

1)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}\)

MTC: \(x\left(x-3\right)\left(x+3\right)\)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{ x\left(3-2x\right)}{x\left(x-3\right)\left(x+3\right)}\dfrac{3x-2x^2}{x\left(x-3\right)\left(x+3\right)}\)

2)

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

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

MTC: \(2x\left(1-x\right)^2\)

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

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

Phần còn lại nhé :v

3.

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

\(x^2-x+1=x^2-x+1\)

\(x+1=x+1\)

MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

4.

\(5x\)

\(x-2y=x-2y=-\left(2y-x\right)\)

\(8y^2-2x^2=2\left(4y^2-x^2\right)=2\left(2y-x\right)\left(2y+x\right)\)

MTC: \(-10x\left(2y-x\right)\left(2y+x\right)\)

\(\dfrac{7}{5x}=\dfrac{7\left(2y-x\right)\left(2y+x\right)-2}{5x\left(2y-x\right)\left(2y+x\right)-2}=\dfrac{-14\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{4}{x-2y}=\dfrac{4\left(2y-x\right)\left(2y+x\right)10x}{-\left(2y-x\right)\left(2y+x\right)10x}=\dfrac{40x\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{\left(x-y\right)-5x}{2\left(2y-x\right)\left(2y+x\right)-5x}=\dfrac{-5x\left(x-y\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

5.

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

\(x^2-x=x\left(x-1\right)\)

\(x^2+x+1\)

MTC: \(x\left(x-1\right)\left(x^2+x+1\right)\)

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

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

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

6.

\(x^2-2ax+a^2=\left(x-a\right)^2\)

\(x^2-ax=x\left(x-a\right)\)

MTC: \(x\left(x-a\right)^2\)

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

\(\dfrac{x+a}{x^2-ax}=\dfrac{\left(x+a\right)\left(x-a\right)}{x\left(x-a\right)\left(x-a\right)}=\dfrac{x^2-a^2}{x\left(x-a\right)^2}\)

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

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

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

b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)

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

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

\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)