a: \(=\dfrac{x^4+15x+7}{x^4+15x+7}\cdot\dfrac{x}{14x^2+1}\cdot\dfrac{4x^3+4}{2x^3+2}=\dfrac{2x}{14x^2+1}\)

b: \(=\dfrac{x^7+3x^2+2}{x^7+3x^2+2}\cdot\dfrac{x^2+x+1}{x^3-1}\cdot\dfrac{3x}{x+1}\)

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

\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(ĐKXĐ:x\ne1\)

\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)

\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)

\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)

\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)

\(\Rightarrow-4x^2-8x\)

⇒-4x(x-4)

Nguyễn Ngọc Thanh Trúc đề là gì

thực hiện phép tính

a) \(\dfrac{25x^3y}{7z}.\dfrac{28z}{15x^2y^5}\)

\(=\dfrac{25x^3y.28z}{7z.15x^2y^5}\)

\(=\dfrac{700x^3yz}{105x^2y^5z}\)

\(=\dfrac{20x}{3y^4}\)

b) \(\dfrac{x^2+3x+9}{2x+10}.\dfrac{x+5}{x^3+27}\)

\(=\dfrac{\left(x^2+3x+9\right)\left(x+5\right)}{\left(2x+10\right)\left(x^3+27\right)}\)

\(=\dfrac{x^3+3x^2+9x+5x^2+15x+45}{2x^4+54x+10x^3+270}\)

\(=\dfrac{x^3+8x^2+24x+45}{2x^4+10x^3+54x+270}\)

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

a) ĐKXĐ: \(x\notin\left\{-1;-2;2\right\}\)

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

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

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

Suy ra: \(x-2-3x-6=-x-1\)

\(\Leftrightarrow-2x-8+x+1=0\)

\(\Leftrightarrow-x-7=0\)

\(\Leftrightarrow-x=7\)

hay x=-7(thỏa ĐK)

Vậy: S={-7}

a) ĐKXĐ: 

Ta có: ⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)

⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)

Suy ra: 

hay x=-7(thỏa ĐK)

Vậy: S={-7}

Đọc tiếp

1, bạn xem lại đề 

2, 15(x-3) + 8x-21 = 12(x+1) +120 

<=> 23x - 66 = 12x + 132 

<=> 11x = 198 <=> x = 198/11 

3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4 

<=> 30x + 10 - 95 = 18x -12

<=> 12x = 73 <=> x = 73/12 

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

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

b.

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

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

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

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

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