Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
Bài 1:
1, \(\frac{2x-5}{x+5}=3\) (ĐKXĐ: x \(\ne\) -5)
\(\Leftrightarrow\) \(\frac{2x-5}{x+5}=\frac{3\left(x+5\right)}{x+5}\)
\(\Rightarrow\) 2x - 5 = 3(x + 5)
\(\Leftrightarrow\) 2x - 5 = 3x + 15
\(\Leftrightarrow\) 2x - 3x = 15 + 5
\(\Leftrightarrow\) -x = 20
\(\Leftrightarrow\) x = -20 (TMĐKXĐ)
Vậy S = {-20}
2, \(\frac{4}{x+1}=\frac{3}{x-2}\) (ĐKXĐ: x \(\ne\) -1; x \(\ne\) 2)
\(\Leftrightarrow\) \(\frac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\) 4(x - 2) = 3(x + 1)
\(\Leftrightarrow\) 4x - 8 = 3x + 3
\(\Leftrightarrow\) 4x - 3x = 3 + 8
\(\Leftrightarrow\) x = 11 (TMĐKXĐ)
Vậy S = {11}
3, \(\frac{5}{2x-3}=\frac{1}{x-4}\) (ĐKXĐ: x \(\ne\) \(\frac{3}{2}\); x \(\ne\) 4)
\(\Leftrightarrow\) \(\frac{5\left(x-4\right)}{\left(2x-3\right)\left(x-4\right)}=\frac{2x-3}{\left(2x-3\right)\left(x-4\right)}\)
\(\Rightarrow\) 5(x - 4) = 2x - 3
\(\Leftrightarrow\) 5x - 20 = 2x - 3
\(\Leftrightarrow\) 5x - 2x = -3 + 20
\(\Leftrightarrow\) 3x = 17
\(\Leftrightarrow\) x = \(\frac{17}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{17}{3}\)}
Bài 2:
1, \(\frac{1}{x-1}+\frac{2}{x+1}=\frac{5x-3}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)
\(\Leftrightarrow\) \(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5x-3}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow\) x + 1 + 2(x - 1) = 5x - 3
\(\Leftrightarrow\) x + 1 + 2x - 2 = 5x - 3
\(\Leftrightarrow\) 3x - 1 = 5x - 3
\(\Leftrightarrow\) 3x - 5x = -3 + 1
\(\Leftrightarrow\) -2x = -2
\(\Leftrightarrow\) x = 1 (KTM)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
2, \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\) (ĐKXĐ: x \(\ne\) 2; x \(\ne\) 0)
\(\Leftrightarrow\) \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow\) x(x + 2) - x + 2 = 2
\(\Leftrightarrow\) x2 + 2x - x + 2 = 2
\(\Leftrightarrow\) x2 + x = 2 - 2
\(\Leftrightarrow\) x2 + x = 0
\(\Leftrightarrow\) x(x + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 0 và x = -1
Ta có: x = 0 KTM đkxđ
\(\Rightarrow\) x = -1
Vậy S = {-1}
3, \(\frac{5}{x-3}-\frac{3}{x+3}=\frac{3x}{x^2-9}\) (ĐKXĐ: x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\) 5(x + 3) - 3(x - 3) = 3x
\(\Leftrightarrow\) 5x + 15 - 3x + 9 = 3x
\(\Leftrightarrow\) 2x + 24 = 3x
\(\Leftrightarrow\) 2x - 3x = 24
\(\Leftrightarrow\) -x = 24
\(\Leftrightarrow\) x = -24 (TMĐKXĐ)
Vậy S = {-24}
Chúc bn học tốt!!
Mình tính mãi vẫn có chỗ sai, mong bạn thông cảm!!
Mình bt mình sai đâu r Garuda
câu 3 bài 3 cuối có cái đoạn 2x + 24 = 3x
\(\Leftrightarrow\) 2x - 3x = -24 (đoạn kia mình ghi là 24 nên quên không đổi dấu)
\(\Leftrightarrow\) -x = -24
\(\Leftrightarrow\) x = 24
Vậy S = {24}
(mình sửa lại rồi nha, chắc hết chỗ sai rồi)
\(\begin{array}{l}a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x - 1}} - \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} - \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} - \frac{2}{{x - 1}}} \right) + \left( {\frac{3}{{x + 2}} - \frac{3}{{x + 2}}} \right)\\ = 0 + \frac{2}{{x + 1}} - \frac{2}{{x - 1}} + 0\\ = \frac{{2\left( {x - 1} \right) - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{2{\rm{x}} - 2 - 2{\rm{x}} - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\end{array}\)
\(\begin{array}{l}b)\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{{x - 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} - 9}} - \frac{3}{{x + 3}}} \right)\\ = 0 + 0 + \frac{3}{{\left( {x + 3} \right)\left( {x - 3} \right)}} - \frac{3}{{x + 3}}\\ = \frac{{3 - 3\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{12 - 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\end{array}\)
1, Đk x≠2;-2
\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=0\\ =>\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{\left(x+2\right)^2}{2\left(x^2-4\right)}-\frac{8x}{2\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x-2}{2\left(x+2\right)}=0\\ =>x-2=0\\ =>x=2\left(loại\right)\)
1) Ta có: x-4=2x+4
\(\Leftrightarrow x-4-2x-4=0\)
\(\Leftrightarrow-x-8=0\)
\(\Leftrightarrow-x=8\)
hay x=-8
Vậy: S={8}
2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)
\(\Leftrightarrow\frac{3\left(2x-1\right)}{6}-\frac{2x}{6}=\frac{6x}{6}-\frac{x}{6}\)
\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)
\(\Leftrightarrow6x-3-2x-6x+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: S={-3}
3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)
Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}-\frac{x\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)
\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)
\(\Leftrightarrow-4x^2-2x-18=0\)
\(\Leftrightarrow-4\left(x^2+\frac{1}{2}x+\frac{4}{5}\right)=0\)
\(\Leftrightarrow x^2+\frac{1}{2}x+\frac{4}{5}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)
Vậy: S=\(\varnothing\)
4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)
\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x-1-24+2x=0\)
\(\Leftrightarrow8x-25=0\)
\(\Leftrightarrow8x=25\)
hay \(x=\frac{25}{8}\)
Vậy: \(S=\left\{\frac{25}{8}\right\}\)
Bài 1:
a/ \(x\ne1;2\)
\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x-2-7x+7+1=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Rightarrow x=1\) (loại)
Vậy pt vô nghiệm
b/ \(x\ne\frac{3}{2}\)
\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)
\(\Leftrightarrow20x+30-15-8x+12=0\)
\(\Leftrightarrow12x+27=0\)
\(\Rightarrow x=-\frac{9}{4}\)
c/ \(x\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Rightarrow x=4\)
Bài 1:
d/\(x\ne\pm3\)
\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
\(\Rightarrow0=0\)
Vậy pt có vô số nghiệm \(x\ne\pm3\)
e/ \(x\ne\pm1\)
\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Leftrightarrow x^2-2x+1+2+3x-3=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)
A= ... (mik k rảh viết vào) 3/2 > ... vì 3/2 > 1 còn ... < 1
B = .... 2/3 < vì 2/3 <1 còn ... > 1