giải các bất phương trình sau:
1) \(\dfrac{x^2-2x+5}{x-2}-x+1\ge0\) 2) \(\dfrac{2x-3}{x+1}-2< 0\)
Bài 1. Giải các bất phương trình sau 1) \(\dfrac{2x-1}{x+1}-2< 0\) 2) \(\dfrac{x^2-2x+5}{x-2}-x+1\ge0\)
3) \(\dfrac{\left(1+2x\right)\left(x-3\right)}{\left(2x+3\right)\left(1-x\right)}\le0\) 4) \(\left|2x-3\right|>5\) 5)\(\left|1-2x\right|\le4\)
6) \(\left|3x+1\right|>x-2\)
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)
\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)
Kết hợp ĐKXĐ.
\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)
\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)
Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)
Ta có bảng sau:
\(x\) | \(-\infty\) \(-\dfrac{3}{2}\) \(-\dfrac{1}{2}\) \(1\) \(2\) \(+\infty\) |
\(1+2x\) | - | - 0 + | + | + |
\(x-2\) | - | - | - | - 0 + |
\(2x+3\) | - 0 + | + | + | + |
\(1-x\) | + | + | + 0 - | - |
\(f\left(x\right)\) | - || + 0 - || + 0 - |
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].
Giải các bất phương trình sau:
1) \(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\)
2) \(\dfrac{\left(3-2x\right)x^2}{\left(x-1\right)}\ge0\)
3) \(\dfrac{2x}{x-1}\le\dfrac{5}{2x-1}\)
1.
ĐK: \(x\ne7;x\ne-1;x\ne3\)
\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)
TH1: \(x< -1\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\ge0\)
\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)
TH2: \(-1< x< 3\)
\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)
\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)
\(\Leftrightarrow-x^2+5x-22\ge0\)
\(\Rightarrow\) vô nghiệm
TH3: \(3< x< 7\)
Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)
\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)
TH4: \(x>7\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\le0\)
\(\Rightarrow\) vô nghiệm
Vậy ...
Các bài kia tương tự, chứ giải ra mệt lắm.
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)
b) \(\dfrac{x+3}{x-2}\le0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow-3\le x< 2\)
d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
Giải các bất phương trình sau:
a) \(\left(x^2+3x-4\right)\left(3-2x\right)< 0\)
\(\dfrac{x^2+3x+4}{x^2-2}\ge0\)
\(\dfrac{x\left(x^2+4x+4\right)}{x^2-1}\ge0\)
b) \(\dfrac{3x-2}{2-x}\le-x\)
c) \(\dfrac{x-3}{x+1}>\dfrac{x+4}{x+2}\)
d) \(\dfrac{x+2}{x-2}-\dfrac{x+3}{x-2}>1\)
e) \(|2x-3|>x+1\)
f) \(|2x-5|\le x+1\)
g) \(x-4-|x^2+3x-4|>0\)
h) \(\left|x^2+4x+3\right|>\left|x^2-4x-5\right|\)
Giải bất phương trình sau: \(\dfrac{1-x^2-2x}{x^2+x-2}\ge0\)
\(\Leftrightarrow\dfrac{-x^2-2x+1}{\left(x+2\right)\left(x-1\right)}>=0\)
=>\(\dfrac{x^2+2x-1}{\left(x+2\right)\left(x-1\right)}< =0\)
TH1: x^2+2x-1>=0 và (x+2)(x-1)<0
=>-2<x<1 và \(\left[{}\begin{matrix}x< =-1-\sqrt{2}\\x>=-1+\sqrt{2}\end{matrix}\right.\)
=>\(-1+\sqrt{2}< =x< 1\)
TH2: x^2+2x-1<=0 và (x+2)(x-1)>0
=>(x>1 hoặc x<-2) và \(-1-\sqrt{2}< =x< =-1+\sqrt{2}\)
=>\(-1-\sqrt{2}< =x< -2\)
Giải các bất phương trình, hệ phương trình
a) \(\dfrac{x^2\left(3x-2\right)\left(x^2-1\right)}{\left(-x^2+2x-3\right)\left(2-x\right)^2}\ge0\)
b) \(\dfrac{x-5}{x-1}>2\)
c) \(2x-\sqrt{x^2-5x-14}< 1\)
d) \(x+\sqrt{x^2-4x-5}< 4\)
e) \(\left\{{}\begin{matrix}\left(4-x\right)\left(x^2-2x-3\right)< 0\\x^2\ge\left(x^2-x-3\right)^2\end{matrix}\right.\)
Giải các phương trình, bất phương trình sau:
1) \(\sqrt{3x+7}-5< 0\)
2) \(\sqrt{-2x-1}-3>0\)
3) \(\dfrac{\sqrt{3x-2}}{6}-3=0\)
4) \(-5\sqrt{-x-2}-1< 0\)
5) \(-\dfrac{2}{3}\sqrt{-3-x}-3>0\)
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
Giải các bất phương trình sau:
a)\(\left(x^2+3x-4\right)\left(3-2x\right)\)<0
b) \(\dfrac{x^2+3x+4}{x^2-2}\ge0\)
c) \(\dfrac{x\left(x^2+4x+4\right)}{x^2-1}\ge0\)
a. TH1:
\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)
Vậy nghiệm của BPT:
\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)