Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2001}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\left(2003-x\right)=0\) (vì \(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))
\(\Leftrightarrow x=2003\).
Vậy tập nghiệm của phương trình là \(S=\left\{2003\right\}\).
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x=0
<=> x=0
Vậy x=0
+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)
\(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x.\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,6\right\}\)
+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)
\(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
\(\frac{x+19}{3}+\frac{x+13}{5}=\frac{x+7}{7}+\frac{x+1}{9}\)
\(=>\frac{x+19}{3}+3+\frac{x+13}{5}+3=\frac{x+7}{7}+3+\frac{x+1}{9}+3\)
\(=>\frac{x+28}{3}+\frac{x+28}{5}=\frac{x+28}{7}+\frac{x+28}{9}\)
\(=>\frac{x+28}{3}+\frac{x+28}{5}-\frac{x+28}{7}-\frac{x+28}{9}=0\)
\(=>\left(x+28\right)\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)=0\)
Do :\(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\ne0\)
\(=>x+28=0\)
\(=>x=-28\)
Vậy nghiệm của phương trình trên là : -28
a) \(\frac{3-2x}{5}>\frac{2-x}{3}\)
<=> \(\frac{3\left(3-2x\right)}{15}>\frac{5\left(2-x\right)}{15}\)
<=> \(9-6x>10-5x\)
<=> 9 - 10 > -5x + 6x
<=> x < -1
Vậy nghiệm của bất phương trình là x < -1
b) \(\frac{x-1}{6}-\frac{x-1}{3}\le\frac{x}{2}\)
<=> \(\frac{x-1-2\left(x-1\right)}{6}\le\frac{3x}{6}\)
<=> \(x-1-2x+2\le3x\)
<=> \(-x+1\le3x\)
<=> \(1\le2x\)
<=> x \(\ge\frac{1}{2}\)
Vậy nghiệm của bất phương trình là x > = 1/2
c) \(\frac{x+1}{3}>\frac{2x-1}{6}-2\)
<=> \(\frac{2\left(x+1\right)}{6}>\frac{2x-1-12}{6}\)
<=> 2x + 1 > 2x - 13
<=> 1 > -13 (luôn đúng)
Vậy nghiệm của bất phương trình luôn đúng với mọi x
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
ĐKXĐ:\(x\ne1\)
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow\frac{x^2+x+1+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow x^2+3x-1=3x^2\)\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(KTMĐK\right)\\x=\frac{1}{2}\left(TMĐK\right)\end{cases}}}\)
Vậy nghiệm của pt là \(x=\frac{1}{2}\)
\(pt\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)
\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)
đến đây là pt bậc 2
Giải phương trình sau:
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4.\)-4
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4\)
\(\Rightarrow\frac{x+2001}{5}+1+\frac{x+1999}{7}+1+\frac{x+1997}{9}+1+\frac{x+1995}{11}+1=0\)
\(\Rightarrow\frac{x+2006}{5}+\frac{x+2006}{7}+\frac{x+2006}{9}+\frac{x+2006}{11}=0\)
\(\Rightarrow\left(x+2006\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)=0\)
\(\Rightarrow x+2006=0\)vì \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}>0\)
\(\Rightarrow x=-2006\)