1. Cho M = \(\left(\frac{5x+2}{x^2-10x}+\frac{5x-2}{x^2+10x}\right)\frac{x^2-100}{x^2+4}\)
tìm x để M có nghĩa
2. Cho N = \(\frac{1}{x^2-2x+1}-\left(\frac{x}{x^2-1}-\frac{1}{x^3-x}\right):x^2-2x+1\)
Tìm TXĐ của N
Giải các phương trình,bất phương trình:
c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
l,\(\left(x^2-2x+1\right)-4=0\)
m,\(4x^2+4x++1=x^2\)
Xin đáy ai giúp mình đi
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
Giải các phương trình sau :
\(a,6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(b,\frac{2\left(x-4\right)}{4}-\frac{3+2x}{10}=x+\frac{1-x}{5}\)
\(c,\frac{2x}{3}+\frac{3x-5}{4}=\frac{3\left(2x-1\right)}{2}-\frac{7}{6}\)
\(d,\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
\(e,\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19
Rút gọn
a)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
b)\(\left\{\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right\}:\frac{4x}{10x-5}\)
c)\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)
\(=\frac{x.\left(x+1\right)}{5.\left(x^2-2x+1\right)}.\frac{5x-5}{3x+3}=\frac{x.\left(x+1\right)}{5.\left(x-1\right)^2}.\frac{5\left(x-1\right)}{3.\left(x+1\right)}=\frac{x}{3.\left(x-1\right)}=\frac{x}{3x-3}\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
Rút gọn
a)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
b)\(\left\{\hept{\begin{cases}2x+1\\2x-1\end{cases}-\frac{2x-1}{2x+1}}\right\}:\frac{4x}{10x-5}\)
c)\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
Thực hiện phép tính:
\(a.\left(x+1\right)\left(2x-3\right)+\left(x+2\right)^2-3x^2-3x\)
\(b.\left(x^2+2x-3\right):\left(x+1\right)+x^2-3x-2\)
\(c.\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
\(d.\frac{x-1}{x^2-5x+4}-\frac{4}{x^2-4x}\)
