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1.ĐK: \(x\ge\dfrac{1}{4}\)
bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)
\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)
\(\Leftrightarrow20x^2-x-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)
2.ĐK: \(-2\le x\le\dfrac{5}{2}\)
bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)
\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)
\(\Leftrightarrow x^2< -x^2+x+6\)
\(\Leftrightarrow-2x^2+x+6>0\)
\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)
3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)
.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)
*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)
*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
a) \(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x=0\)
b) \(2x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x\left(x-1\right)+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}-x\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x^2+x}{x-1}\)
\(\Leftrightarrow x^2-x+3=3x-x^2+x\) ( điều kiện \(x\ne1\) )
\(\Leftrightarrow2x^2-5x+3=0\)
\(\Delta=b^2-4ac\)
\(\Delta=1\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\)
c) \(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}\) ( điều kiện \(x>2\) )
\(\Leftrightarrow x^2-4x-2=x-2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\end{matrix}\right.\)
Vậy \(x=5\)
d) \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}\)
\(\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}\) ( điều kiện \(x>\dfrac{3}{2}\) )
\(\Leftrightarrow2x^2-x-3=2x-3\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
$a)\frac{2x}{2x^{2}-5x+3}+\frac{13x}{2x^{2}+x+3}=6$ (1)
Nhận thấy x=0 ko phải nghiệm của phương trình
Chia cả tử và mẫu của mỗi phân thức cho x, ta được:
$\frac{2}{2x-5+\frac{3}{x}}+\frac{13}{2x+1+\frac{3}{x}}=6$
Đặt $2x+\frac{3}{x}$=t
=> (1) <=> $\frac{2}{t-5}+\frac{13}{t+1}=6$
<=> $2t^{2}-13t+11=0$
Có a+b+c=2-13+11=0
=> $t_{1}=1$
$t_{2}=\frac{c}{a}=\frac{11}{2}$
* t = 1
=> $2x+\frac{3}{x}=1$
<=> $2x^{2}-x+3=0$ (vô nghiệm)
* t = $\frac{11}{2}$
=> $2x+\frac{3}{x}=\frac{11}{2}$
<=> $4x^{2}-11x+6=0$
=> $x_{1}=\frac{3}{4}$
$x_{2}=2$
Vậy phương trình có tập nghiệm S={$\frac{3}{4};2$}
b, \(x^2+\left(\dfrac{x}{x-1}\right)^2=1\)
\(\Leftrightarrow\left[x^2+\left(\dfrac{x}{x-1}\right)^2+2.x.\dfrac{x}{x-1}\right]-2.\dfrac{x^2}{x-1}-1=0\)
\(\Leftrightarrow\left(x+\dfrac{x}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\)
\(\Leftrightarrow\left(\dfrac{x\left(x-1\right)+x}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\) (1)
Đặt : \(\dfrac{x^2}{x-1}=t\) (*) thì phương trình (1) trở thành:
\(t^2-2t-1=0\)
Ta có: \(\Delta=8>0\)
\(\Rightarrow t_1=\dfrac{2-\sqrt{8}}{2}=\dfrac{2-2\sqrt{2}}{2}=1-\sqrt{2}\)
\(t_2=\dfrac{2+\sqrt{8}}{2}=\dfrac{2+2\sqrt{2}}{2}=1+\sqrt{2}\)
Thay vào (*) rồi tìm x là xong
=.= hk tốt!!