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ĐK: \(x\ne-5\)
\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
\(\Leftrightarrow x^2+\dfrac{25x^2}{\left(x+5\right)^2}-\dfrac{10x^2}{x+5}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\left(x-\dfrac{5x}{x+5}\right)^2+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\dfrac{x^4}{\left(x+5\right)^2}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow y^2+10y-11=0\left(y=\dfrac{x^2}{x+5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-11\end{matrix}\right.\)
TH1: \(y=1\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=1\)
\(\Leftrightarrow x^2=x+5\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{21}}{2}\left(tm\right)\)
TH2: \(y=-11\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=-11\)
\(\Leftrightarrow x^2=-11x-55\)
\(\Rightarrow\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{1\pm\sqrt{21}}{2}\)
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
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].
TH1: \(x\le-2\)
\(pt\Leftrightarrow-x-1-x-2-2x+1=3\)
\(\Leftrightarrow0x=5\)
\(\Rightarrow\) vô nghiệm
TH2: \(-2< x\le-1\)
\(pt\Leftrightarrow-x-1+x+2-2x+1=3\)
\(\Leftrightarrow x=-\dfrac{1}{2}\left(l\right)\)
TH3: \(-1< x\le\dfrac{1}{2}\)
\(pt\Leftrightarrow x+1+x+2-2x+1=3\)
\(\Leftrightarrow0x=-1\)
\(\Rightarrow\) vô nghiệm
TH4: \(x>\dfrac{1}{2}\)
\(pt\Leftrightarrow x+1+x+2+2x-1=3\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(l\right)\)
Vậy phương trình đã cho vô nghiệm
1: \(\Leftrightarrow\left[{}\begin{matrix}2x-3>5\\2x-3< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\)
2: \(\Leftrightarrow-4< =2x-1< =4\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1>=-4\\2x-1< =4\end{matrix}\right.\Leftrightarrow\dfrac{-3}{2}< =x< =\dfrac{5}{2}\)
4: =>2x-3>5 hoặc 2x-3<-5
=>x>4 hoặc x<-1
5: =>-4<=2x-1<=4
=>-3/2<=x<=5/2
a, ĐKXĐ : \(D=R\)
BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)
Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)
BPTTT : \(5\sqrt{a+24}>a\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)
\(\Leftrightarrow-24\le a< 40\)
- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)
\(\Leftrightarrow-9< x< 4\)
Vậy ....
b, ĐKXĐ : \(x>0\)
BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)
- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)
\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)
BPTTT : \(2a\le a^2\)
\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)
\(\Leftrightarrow a\ge2\)
\(\Leftrightarrow a^2\ge4\)
- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)
\(\Leftrightarrow4x^2-12x+1\ge0\)
\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)
Vậy ...
\(\left(x^2+25+150\right)\left(x^2+30x+216\right)=2x^2\)
\(\Rightarrow\left[\left(x+12,5\right)^2-6,25\right]\left[\left(x+15\right)^2-9\right]=2x^2\)
\(\Rightarrow\left(x+15\right)\left(x+10\right)\left(x+18\right)\left(x+12\right)=2x^2\)
Đến đây tách như lớp 8 ,dài quá nên mk lười :) bạn tự giải nha
thanks làm tiếp để thế hệ sau tham khảo :
\(\Rightarrow\left(x+15\right)\left(x+12\right)\left(x+10\right)\left(x+18\right)=2x^2\)
\(\Rightarrow\left(x^2+27x+180\right)\left(x^2+28x+180\right)=2x^2\)
Chia cả hai vế cho x2
\(\Rightarrow\left(x+27+\frac{180}{x}\right)\left(x+28+\frac{180}{x}\right)=2\)
Đặt \(a=x+\frac{180}{x}\)
\(\Rightarrow\left(a+27\right)\left(a+28\right)=2\)
\(\Rightarrow a^2+56a+756=2\)
\(\Rightarrow a^2+56a+754=0\)
tìm nghiệm rồi thế vào :