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

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

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

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

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) \(\frac{3x-2}{2xy}-\frac{7x-4}{2xy}\)

\(=\frac{3x-2}{2xy}+\frac{-\left(7x-4\right)}{2xy}\)

\(=\frac{3x-2-7x+4}{2xy}\)

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

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

\(=\frac{-2x+1}{xy}.\)

b) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)

Ta có:

\(x^2+4x=x.\left(x+4\right)\)

\(2x+8=2.\left(x+4\right)\)

\(MTC:2x.\left(x+4\right)\)

\(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)

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

\(=\frac{6.2}{2x.\left(x+4\right)}+\frac{3x}{2x.\left(x+4\right)}\)

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

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

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

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

\(=\frac{3}{2x}.\)

Chúc bạn học tốt!

an có 10000000 quả cam an cho mẹ gấp đôi rồi an co ba số quả lớn hơn mẹ 200 vậy an còn bao nhiêu quả cam

a) \(\frac{5x-1}{3x^2y}+\frac{x-1}{3x^2y}=\frac{5x-1+x-1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)

b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7\left(\frac{3}{2}x^2\right)}{18x^3y^2}+\frac{11y}{18x^3y^2}=\frac{10,5x^2+11y}{18x^3y^2}\)

c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)}{\left(x+2\right)\left(4x-7\right)}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)

a) \(\frac{5x-1}{3x^2y}+\frac{x+1}{3x^2y}=\frac{5x-1+x+1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)

b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7x^2.18+11.12y}{12x^3y^2.18}=\frac{126x^2+132y}{216x^3y^2}=\frac{6\left(21x^2+22y\right)}{216x^3y^2}=\frac{21x^2+22y}{36x^3y^2}\)

c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x-2\right)}{4x-7}\)

a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

Bài 1 :

a, Ta có : \(3x-1=2x+4\)

=> \(3x-2x=4+1\)

=> \(x=5\)

Vậy phương trình có tập nghiệm \(S=\left\{5\right\}\)

b, Ta có : \(5x-2=0\)

=> \(5x=2\)

=> \(x=\frac{2}{5}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{2}{5}\right\}\)

c, Ta có : \(7x-4=3x+12\)

=> \(7x-3x=12+4\)

=> \(4x=16\)

=> \(x=4\)

Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)

d, Ta có : \(\frac{x-1}{2}+\frac{3x+2}{4}=\frac{x-7}{12}\)

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

=> \(6\left(x-1\right)+3\left(3x+2\right)=x-7\)

=> \(6x-6+9x+6=x-7\)

=> \(6x+9x-x=6-7-6\)

=> \(14x=-7\)

=> \(x=-\frac{1}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{-\frac{1}{2}\right\}\)

Bài 2 :

a, ĐKXĐ : \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x-1\ne0\end{matrix}\right.\)

=> \(x-1\ne0\)

=> \(x\ne1\)

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

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

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

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

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

= ​​\(\frac{x-2}{x-1}\)

Bài 5 :

a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

=> \(36x+3=0\)

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

Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)

b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)

=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)

=> \(35x-5+60x-96+6x=0\)

=> \(101x-101=0\)

=> \(x=1\)

Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)

c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

=> \(\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(2x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)

=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)

=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)

=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

=> \(-64x+123=0\)

=> \(x=\frac{123}{64}\)

Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)