Giải pt:
\(\frac{2x-3}{x-1}\)=4
giải PT \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{x^2+2x-3}\)
\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\) \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)
\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
Giải PT;
\(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{3}{x}\)
Giải pt : \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{x^2+2x-3}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1\right)^2-4}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1+2\right)\left(x+1-2\right)}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+3\right)\left(x-1\right)}=1\)
ĐKXĐ: x \(\ne\) 1 và x \(\ne\) - 3
\(\left(3x-1\right)\left(x+3\right)-\left(2x-5\right)\left(x-1\right)+4=\left(x+3\right)\left(x-1\right)\)
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 = x2 - x + 3x - 3
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 - x2 + x - 3x + 3 = 0
13x - 1 = 0
x = \(\frac{1}{13}\)
giải pt sau: \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
\(\frac{1}{x^2-2x+2}-1+\frac{2}{x^2-2x+3}-1+2-\frac{6}{x^2-2x+4}=0\)
\(\Leftrightarrow\frac{-x^2+2x-1}{x^2-2x+2}+\frac{-x^2+2x-1}{x^2-2x+3}+\frac{2\left(x^2-2x+1\right)}{x^2-2x+4}=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\Rightarrow x=1\\\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:
\(\frac{2}{a+1}-\frac{1}{a-1}-\frac{1}{a}=0\Leftrightarrow\frac{2\left(a-1\right)-\left(a+1\right)}{\left(a^2-1\right)}-\frac{1}{a}=0\)
\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)
\(\Rightarrow x^2-2x+3=\frac{1}{3}\Leftrightarrow x^2-2x+1+\frac{5}{3}=0\)
\(\Leftrightarrow\left(x-1\right)^2+\frac{5}{3}=0\) (vô nghiệm)
Vậy \(x=1\)
Giải pt:
1. x-4=2x+4
2. \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)
3.\(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
4.\(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)
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: \(S=\left\{\frac{25}{8}\right\}\)
giải pt: a. (x - 2)(x+1)(x+3) = (x+3)(x+1)(2x-5)
b. \(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
Giải pt \(\sqrt{1+\frac{2}{x}}=-2x-4+\frac{3}{x}\)
\(\Leftrightarrow\sqrt{\frac{x+2}{x}}=-2x-4+\frac{3}{x}\)
- Với \(x>0\), đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{\frac{1}{x}}=b\end{matrix}\right.\)
\(\Rightarrow ab=-2a^2+3b^2\)
\(\Leftrightarrow2a^2+ab-3b^2=0\Leftrightarrow\left(a-b\right)\left(2a+3b\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x+2=\frac{1}{x}\Leftrightarrow x^2+2x-1=0\)
- Với \(x\le-2\), đặt \(\left\{{}\begin{matrix}\sqrt{-x-2}=a\\\sqrt{-\frac{1}{x}}=b\end{matrix}\right.\)
\(\Rightarrow ab=2a^2-3b^2\Leftrightarrow2a^2-ab-3b^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(2a-3b\right)=0\Leftrightarrow2a=3b\)
\(\Leftrightarrow4\left(-x-2\right)=-\frac{9}{x}\Leftrightarrow...\)
giải pt
\(\frac{1}{\left(x^2+2x+2\right)^2}+\frac{1}{\left(x^2+2x+3\right)^2}=\frac{5}{4}\)
Giải pt
\(\frac{x^2+x+1}{x+1}+\frac{x^2+2x+2}{x+2}=\frac{x^2+3x+3}{x+3}+\frac{x^2+4x+4}{x+4}\)
\(x\ne\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)
\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)
\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)
\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{-2}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)
giải PT:\(\frac{x^4}{2x^2+1}+\frac{2x^2+1}{x^4}=2\)
lm trên cymath.com
ĐK: x khác 0
Đặt: \(\frac{x^4}{2x^2+1}=t>0\Rightarrow\frac{2x^2+1}{x^4}=\frac{1}{t}\)
Ta có phương trình: \(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow\left(t-1\right)^2=0\Leftrightarrow t=1\)
Với t = 1 ta có: \(\frac{x^4}{2x^2+1}=1\)<=> \(x^4-2x^2-1=0\Leftrightarrow\orbr{\begin{cases}x^2=1+\sqrt{2}\\x^2=1-\sqrt{2}\left(loai\right)\end{cases}}\)
khi đó: \(x=\pm\sqrt{1+\sqrt{2}}\)tm
Vậy....
Một cách làm khác :
\(ĐKXĐ:x\ne0\)
Với điều kiện trên thì : \(\hept{\begin{cases}x^4>0\\2x^2+1>0\end{cases}}\)
Do đó áp dụng BĐT AM - GM cho 2 số dương ta có :
\(\frac{x^4}{2x^2+1}+\frac{2x^2+1}{x^4}\ge2.\sqrt{\frac{x^4}{2x^2+1}\cdot\frac{2x^2+1}{x^4}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x^4}{2x^2+1}=\frac{2x^2+1}{x^4}\)
\(\Leftrightarrow x^4=2x^2+1\)
\(\Leftrightarrow x^4-2x^2+1=2\)
\(\Leftrightarrow\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=\sqrt{2}\\x^2-1=-\sqrt{2}\end{cases}}\)
\(\Leftrightarrow x^2-1=\sqrt{2}\Leftrightarrow x^2=\sqrt{2}+1\)
\(\Leftrightarrow x=\pm\sqrt{\sqrt{2}+1}\)( Thỏa mãn ĐKXĐ )
Vậy pt đã cho có tập nghiệm \(S=\left\{\pm\sqrt{\sqrt{2}+1}\right\}\)