Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
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}
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
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: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)
\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)
=>3x-5<=30x-100
=>30x-100>3x-5
=>27x>95
hay x>95/27
b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)
=>26x-8<-11x
=>37x<8
hay x<8/37
a: \(=\dfrac{1}{x+2y}+\dfrac{1}{x-2y}-\dfrac{4y}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{x-2y+x+2y-4y}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2}{x+2y}\)
b: \(=\dfrac{2x}{x-1}+\dfrac{5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}\cdot\dfrac{2\left(x+1\right)}{5\left(1-x\right)}\)
\(=\dfrac{2x}{x-1}-2=\dfrac{2x-2x+2}{x-1}=\dfrac{2}{x-1}\)
c: \(=\dfrac{5\left(x-1\right)}{2x}\cdot\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{5\cdot4x}{2x\cdot\left(x+1\right)}=\dfrac{10}{x+1}\)