\(M=\dfrac{x-3\sqrt{x}+2+x+3\sqrt{x}+2-2x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\sqrt{x}-2}{3\sqrt{x}-6}\)

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

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

a) (2x - 1)(3x + 5) - 2(-4x + 1)² = 6x² + 10x - 3x - 5 - 2(16x² - 8x + 1) = 6x² - 3x - 5 - 32x² + 16x - 2 = -26x² + 13x - 7

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

c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)

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

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

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

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

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

d) (x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

= (x - 1 - x - 1)[(x - 1)² + (x - 1)(x + 1) + (x + 1)²] + 6(x² - 1)

= -2(x² - 2x + 1  + x² - 1 + x² + 2x + 1) + 6x² - 6

= -2(3x² + 1) + 6x² - 6

= -6x² - 2 + 6x²  - 6

= -8

e) (2x + 7)² - (4x + 14)(2x - 8) + (8 - 2x)²

= (2x + 7)² - 2(2x + 7)(2x - 8) + (2x - 8)²

= (2x + 7 - 2x + 8)²

= 15² = 225

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

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}

Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)

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

ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)

\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)

Thỏa mãn ĐK

Các trường hợp khác làm tương tự

\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

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

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)

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