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`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}`
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 ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x}{x+2y}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
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: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)
\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{3x^2+3x+x+1-\left(x^2-2x+1\right)-\left(x^2-x+3x-3\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\frac{x^2+4x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}\)
\(=\frac{x+3}{x^2-2x+1}\)
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
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 !
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: \(\left(\frac{x+1}{x-3}+\frac{5x-39}{x^2-9}-\frac{11}{x+3}\right):\frac{x^2+2x+1}{2x+6}\)
\(=\left(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{5x-39}{\left(x-3\right)\left(x+3\right)}-\frac{11\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2+4x+3+5x-39-11x+33}{\left(x+3\right)\left(x-3\right)}\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2-2x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x^2-3x+x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{\left(x-3\right)\left(x+1\right)\cdot2}{\left(x-3\right)\left(x+1\right)^2}\)
\(=\frac{2}{x+1}\)