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Bài 1:
a) Ta có: \(\left(12x^3-28x^2+21x-5\right):\left(6x-5\right)-\left(2x^2-4x\right)\)
\(=\left(12x^3-10x^2-18x^2+15x+6x-5\right):\left(6x-5\right)-\left(2x^2-4x\right)\)
\(=\frac{2x^2\left(6x-5\right)-3x\left(6x-5\right)+\left(6x-5\right)}{6x-5}-2x^2+4x\)
\(=\frac{\left(6x-5\right)\left(2x^2-3x+1\right)}{6x-5}-2x^2+4x\)
\(=2x^2-3x+1-2x^2+4x\)
\(=x+1\)
b) Ta có: \(\left(\frac{x+1}{x-3}+\frac{5x-39}{x^2-9}-\frac{11}{x+3}\right):\frac{x^2+2x+1}{2x+6}\)
\(=\left(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{5x-39}{\left(x-3\right)\left(x+3\right)}-\frac{11\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2+4x+3+5x-39-11x+33}{\left(x+3\right)\left(x-3\right)}\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2-2x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x^2-3x+x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{\left(x-3\right)\left(x+1\right)\cdot2}{\left(x-3\right)\left(x+1\right)^2}\)
\(=\frac{2}{x+1}\)
a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)
b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
Bài 1: Thực hiện phép tính
a) Ta có: \(3x^2\left(5x^2-2x+4\right)\)
\(=15x^4-6x^3+12x^2\)
b) Ta có: \(\left(2x^2-4\right)\left(x^2-3\right)\)
\(=2x^4-6x^2-4x^2+12\)
\(=2x^4-10x^2+12\)
c) Ta có: \(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\cdot\left(1-\frac{1}{x^2}\right)\)
\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{1-x^2}{x^2}\)
\(=\frac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{-\left(x-1\right)\left(x+1\right)}{x^2}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{-x^2}\)
\(=\frac{4x}{-x^2}=\frac{-4x}{x^2}=\frac{-4}{x}\)
d) Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)
\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{3x^2+3x+x+1-\left(x^2-2x+1\right)-\left(x^2-x+3x-3\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{x^2+4x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}\)
\(=\frac{x+3}{x^2-2x+1}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
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)
$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$
$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$
$=\frac{5x}{6(x-1)}$
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)
$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$
$=\frac{-3(x+7)}{2x+1}$
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
Giải tiêu biểu câu a nhé.
a/ \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow x=-\frac{5}{19}\)
a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)
c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)
d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)
`1a,(5x + 3)(4x^2 - 6x + 2)`
`=5x(4x^2-6x+2)+3(4x^2-6x+2)`
`=20x^3-30x^2+10x+12x^2-18x+6`
`=20x^3(-30x^2+12x^2)+(10x-18x)+6`
`=20x^3-18x^2-8x+6`
`b,x/(2x-2)+(x-2)/(x^2-1)-5/(2x+2)` `Đkxđ: x ≠ +-1`
`<=>x/(2(x-1))+(x-2)/((x+1)(x-1))-5/(2(x+1))`
`<=>(x(x+1)+2(x-2)-5(x-1))/(2(x-1)(x+1))`
`<=>(x^2-2x+1)/(2(x-1)(x+1))`
`<=>(x-1)/(2(x+1))`
`2,(x+3)(x-2)-(x^3-3x^2+3x-1):(x-1)=2`
`<=>(x+3)(x-2)-(x^3-3x^2+3x-1)/(x-1)=2`
`<=>(x+3)(x-2)-(x-1)^2=2`
`<=>x^2-2x+3x-6-x^2+2x-1=2`
`<=>3x-7=2`
`<=>x=3`
Vậy `S={3}`