Các ĐKXĐ: bạn tự tìm

a)

\(\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}=\frac{11x+10}{3(x-1)}-\frac{15x+13}{4(x-1)}=\frac{4(11x+10)-3(15x+13)}{12(x-1)}\)

\(=\frac{-x+1}{12(x-1)}=\frac{-(x-1)}{12(x-1)}=\frac{-1}{12}\)

b)

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

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

c)

\(\frac{4xy-1}{5x^2y}-\frac{2xy-1}{5x^2y}=\frac{(4xy-1)-(2xy-1)}{5x^2y}=\frac{2xy}{5x^2y}=\frac{2}{5x}\)

d)

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

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

$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$

$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)

$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$

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

f)

$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$

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

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

$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$

$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$

h)

$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$

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

$=\frac{5x}{6(x-1)}$

ĐKXĐ bạn tự tìm nha : )

k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)

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

j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)

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

i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)

\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)

h, = k,

f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

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

b. \(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{12x.15y^4}{5y^3.8x^3}=\frac{3.3y}{2x^2}=\frac{9y}{2x^2}\)

c. \(\frac{4y^2}{11x^4}.\left(\frac{-3x^2}{8y}\right)=\frac{4y^2.\left(-3x^2\right)}{11x^4.8y}=\frac{-3y}{22x^2}\)

d. \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}=\frac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\frac{x+2}{6}\)

f. \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x+6\right)\left(x-6\right).3}{\left(2x+10\right)\left(6-x\right)}=\frac{-3x-18}{2x+10}\)

g. \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6}=\frac{\left(x^2-9y^2\right).3xy}{x^2y^2.\left(2x-6\right)}=\frac{3x^2-27y^2}{2x^2y-6xy}\)

h. \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right).3x}{x\left(x+4\right).2\left(1-2x\right)}=\frac{3+6x}{2x+8}\)

i. \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right).2\left(a-b\right)\left(a+b\right)}{\left(b-a\right)\left(a+b\right)}=-2a^2-2ab\)

j. \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{\left(x+y\right).3\left(x-y\right)\left(x+y\right)}{\left(y-x\right).x\left(x+y\right)}=\frac{-3x-3y}{x}\)

Bài làm

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

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

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

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

b) Bạn kiểm tra lại đề bài

c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)

\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)

\(\Leftrightarrow\frac{-3}{4x+1}-\frac{2}{4x+1}+\frac{8}{\left(4x+1\right)\left(4x-1\right)}=0\)

\(\Leftrightarrow\frac{-3\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3}{\left(4x-1\right)\left(4x+1\right)}-\frac{8x-2}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3-8x+2+8}{\left(4x-1\right)\left(4x+1\right)}=0\)

=> -20x+13=0

<=> -20x=-13

<=> \(x=\frac{13}{20}\left(tmđk\right)\)

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}

) \(\dfrac{x^3+8y^3}{2y+x}\)

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

\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)

\(=x^2+2xy+4y^2\)

b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)

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

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

\(=\dfrac{3a-1}{2\left(a-4\right)}\)

c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)

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

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

d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)

\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)

\(=x^2-10x+25+7x+14-x^2-2x\)

\(=39-5x\)

e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)

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

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

\(=\dfrac{5x+1}{x-2}\)

h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)

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

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

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

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

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

\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)

\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)

=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0

=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0

=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0

=>(2x^2+120+35x)(2x^2+31x+120)=0

=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)

b: Đặt x^2-3x=a

Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)

\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)

=>(3a+10)(a+5)=6(a^2+7a+12)

=>6a^2+42a+72=3a^2+15a+10a+50

=>3a^2+17a+22=0

=>x=-2 hoặc x=-11/3