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\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+5\right)\left(5x+5\right)}-\dfrac{x^2-25}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(3x-3-x\right)\left(3x-3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{x^2-25}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{3\left(x+5\right)\cdot5\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(x+5\right)}{5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2-\left(x+5\right)\left(x+1\right)}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-6x-5}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3x^2+18x+31}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3\left(x+3\right)\left(3x^2+18x+31\right)-\left(2x-3\right)\left(4x-3\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{\left(3x+9\right)\left(3x^2+18x+31\right)-\left(8x^2-18x+9\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3+8x^2-18x^2-18x+9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3-10x^2-9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{x^3+91x^2+264x+270}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
1) \(\dfrac{A\left(x-5\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{3x\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow A=3x\)
2) \(\dfrac{\left(x+3\right)\left(x-2\right)}{A\left(x-3\right)}=\dfrac{\left(5x-1\right)\left(x-2\right)}{\left(5x-1\right)\left(x^2+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)}{A\left(x-3\right)}=\dfrac{1}{\left(x^2+3\right)}\)
\(\Rightarrow A=\dfrac{\left(x^2+3\right)\left(x+3\right)}{x-3}\)
3) \(\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x+5\right)\left(2x-3\right)}=\dfrac{\left(x-5\right)A}{\left(2x-3\right)\left(x+2\right)}\)
\(\Leftrightarrow1=\dfrac{A}{\left(x+2\right)}\)
\(\Leftrightarrow A=x+2\)
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)