a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

sorry em mới lớp 7

lớp 7 kệ e chị có bắt e tl đâu

a: \(=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

b: \(=\dfrac{6x+3\left(x-1\right)+2\left(x-2\right)}{6}=\dfrac{6x+3x-3+2x-4}{6}=\dfrac{11x-7}{6}\)

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

\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)

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

ĐK: x≠1

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

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

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

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x²-2x+1)=0

<=> -(6x²-x+12x-2)+9x²+4x-18x-8-3x²+2x-1 = 0

<=> -6x²-11x+2+9x²+4x-18x-8-3x²+2x-1=0

<=> -23x-7=0

<=> -23x=7

<=> x= \(\frac{-7}{23}\left(tm\right)\)

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)

⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

⇒x²+x+1+2x²-5=4x-4

⇔3x²-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}

\(a,\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)\(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow x^2+x-x^2+2x-1+x-3+x+1=0\)

\(\Leftrightarrow5x-3=0\)

\(\Leftrightarrow x=\frac{3}{5}\left(tmđk\right)\)

Vậy ......

\(b,\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\) \(Đkxđ:....\)

\(\Leftrightarrow\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow-4x+8-16=10\)

\(\Leftrightarrow x=-\frac{9}{2}\)

Vậy ...............

b, ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x+3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne-3\end{matrix}\right.\)

- Ta có : \(\frac{x-3}{x+1}-\frac{x+1}{x+3}=\frac{x^2-x-10}{x^2+4x+3}\)

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

=> \(\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x+1\right)=x^2-x-10\)

=> \(x^2-9-x^2-2x-1-x^2+x+10=0\)

=> \(-x-x^2=0\)

=> \(x\left(x+1\right)=0\)

=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

=> x = 0 .

Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ne-1\end{matrix}\right.\)

Ta có : \(\frac{x}{2x-6}+\frac{x}{2x+2}+\frac{2x}{\left(x+1\right)\left(3-x\right)}=0\)

=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{4x}{\left(x+1\right)\left(x-3\right)}=0\)

=> \(x\left(x+1\right)+x\left(x-3\right)-4x=0\)

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

=> \(x\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)

\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)

Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)

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

\(\Leftrightarrow4x-2-6x-3=4\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)

Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)

\(b,ĐKXĐ:x\ne\pm1;-3\)

Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)

\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)

\(\Leftrightarrow9x^2+14x+13=0\)

\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)

\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)

Pt vô nghiệm 

\(c,ĐKXĐ:x\ne1\)

Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\pm1\)

Kết hợp vs ĐKXĐ được x = -1

Vậy pt có nghiệm duy nhất x = -1

làm lần lượt nha(bài nào k bt bỏ qua)

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

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

\(\Rightarrow-2x-5=4\)

\(\Rightarrow-2x=9\)

\(\Rightarrow x=\frac{9}{-2}\)

a)\(ĐKXĐ:x\ne0;-1\)

Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

a) Điều kiện xác định: x khác 0, x khác 1/2

(x+3)(2x-1)/x(2x-1) = (2x+2)x/(2x-1)x

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

2x² -x+6x-3 = 2x² +2x

2x² -x+6x-3-2x² -2x = 0

3x-3=0

3x=3

x=1 ( thỏa mãn đièu kiện xác định)

Vậy phương trình dã cho có tập nghiệm là S=[1]

b) Điều kiện xác định: x khác -1

5x/2x+2 + 2x+2/2x+2 = -12/2x+2

5x+2x+2/2x+2 = -12/2x+2

7x+2/2x+2 = -12/2x+2

7x+2=-12

7x=-14

x=-2 ( thỏa mãn đièu kiện xác định)

Vậy phương trình đã cho có tập nghiệm là S=[-2]

c) Điều kiện xác định: x khác -1, x khác 0

(x+3)x/(x+1)x + (x-2)(x+1)/x(x+1) = 2x(x+1)/x(x+1)

(x+3)x+(x-2)(x+1)/x(x+1) = 2x(x+1)/x(x+1)

(x+3)x+(x-2)(x+1) = 2x(x+1)

x² +3x+(x² +x-2x-2)=2x² +2x

x² +3x+(x² -x-2) = 2x² +2x

x² +3x+x² -x-2=2x² +2x

2x² +2x-2=2x² +2x

2x² +2x-2-2x² -2x=0

0x-2=0 ( vô lý)

Vậy phương trình này vô nghiệm 

bài này thực ra mk làm ở trên lớp rùi nên mk mới trả lời 

hok tot

mk quên ko nói: bạn cho dấu tương đương vào trước mỗi dòng giải phương trình nhé

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

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

\(\Leftrightarrow2x^2+x-1=2x^2+2x\)

\(\Leftrightarrow x-1=2x\)

\(\Leftrightarrow x=-1\)

b)\(\frac{5x}{2x+2}+1=\frac{-6}{x+1}\)

\(\Leftrightarrow\frac{5x+2x+2}{2\left(x+1\right)}=\frac{-12}{2\left(x+1\right)}\)

\(\Leftrightarrow7x+2=-12\)

\(\Leftrightarrow x=-2\)

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

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

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

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

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

\(\Leftrightarrow x=-3\)