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lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0
x+√x>(2√x+3)(√x−1)
⇔x+√x>2x+√x−3
⇔x−3>0
⇔x>3. (tmđk).
Điều kiện \(x^2-2x\ge0\Leftrightarrow\left[\begin{array}{nghiempt}x\ge2\\x\le0\end{array}\right.\) khi đó :
Bất phương trình \(\Leftrightarrow3^{\sqrt{x^2-2x}}\ge\left(3\right)^{\sqrt{\left(x-1\right)^2}-x}\Leftrightarrow\sqrt{x^2-2x}\ge\left|x-1\right|-x\)
- Khi \(x\ge2\Rightarrow x-1>0\) nên bất phương trình \(\sqrt{x^2-2x}\ge-1\) đúng với mọi \(x\ge2\)
- Khi \(x\le0\Rightarrow x-1< 0\) nên bất phương trình \(\sqrt{x^2-2x}\ge1-2x\)
\(\Leftrightarrow\begin{cases}x^2-2x\ge1-4x+4x^2\\x\le0\end{cases}\) vô nghiệm
Vậy tập nghiệm của bất phương trình là : S = [2;\(+\infty\) )
Câu 1:
a/ \(x\ge-11\)
Đặt \(\sqrt{x+11}=a\ge0\Rightarrow11=a^2-x\), pt đã cho trở thành:
\(x^2+a=a^2-x\Leftrightarrow x^2-a^2+x+a=0\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)
TH1: \(x+a=0\Leftrightarrow x+\sqrt{x+11}=0\Leftrightarrow-x=\sqrt{x+11}\)
\(\Leftrightarrow\left[{}\begin{matrix}-x\ge0\\x^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2-x-11=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-3\sqrt{5}}{2}\)
TH2: \(x-a+1=0\Leftrightarrow x+1=\sqrt{x+11}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\\left(x+1\right)^2=x+11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-10=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{41}}{2}\)
b/ \(\sqrt{9+x}=x-9\Leftrightarrow\left\{{}\begin{matrix}x-9\ge0\\9+x=\left(x-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge9\\x^2-19x+72=0\end{matrix}\right.\) \(\Rightarrow x=\frac{19+\sqrt{73}}{2}\)
Câu 2:
a/
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-1\right)\left(x-4\right)}=\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-4\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -1\\x>4\\1< x< 3\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-1< x< 1\\3< x< 4\end{matrix}\right.\)
\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
\(f\left(x\right)\) ko xác định tại \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b/ \(h\left(x\right)=\frac{-x^2+3x-1}{\left(x^2-2x+3\right)\left(x+2\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -2\\\frac{3-\sqrt{5}}{2}< x< \frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-2< x< \frac{3-\sqrt{5}}{2}\\x>\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)=0\) tại \(x=\frac{3\pm\sqrt{5}}{2}\)
\(f\left(x\right)\) ko xác định tại \(x=-2\)
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
1. \(\Leftrightarrow\left(3x-1\right)\left(\sqrt{5}x-2\right)\ge0\Rightarrow\left[{}\begin{matrix}x\le\frac{1}{3}\\x\ge\frac{2}{\sqrt{5}}\end{matrix}\right.\)
2. \(\Leftrightarrow\frac{\left(3-2x\right)\left(3+2x\right)}{2x-3}\ge0\Leftrightarrow\left[{}\begin{matrix}x\ne\frac{3}{2}\\x\le-\frac{3}{2}\end{matrix}\right.\)
3. \(\left|x-2\right|\ge3\Leftrightarrow\left[{}\begin{matrix}x-2\ge3\\x-2\le-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)
4. \(\Leftrightarrow-10\le3x+1\le10\Rightarrow-\frac{11}{3}\le x\le3\)
5. \(\Leftrightarrow\frac{3x^2-x+2}{x^2-9}-3\le0\Leftrightarrow\frac{-x+29}{\left(x-3\right)\left(x+3\right)}\le0\Rightarrow\left[{}\begin{matrix}-3< x< 3\\x\ge29\end{matrix}\right.\)
6. \(\Leftrightarrow\frac{4}{\left(x-2\right)^2}+\frac{1}{x-2}>0\Leftrightarrow\frac{x+2}{\left(x-2\right)^2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-2\\x\ne2\end{matrix}\right.\)
a/ \(\Leftrightarrow\left(4-x\right)\left(x+1\right)\left(x-8\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}x< -1\\4< x< 8\end{matrix}\right.\)
b/ \(\frac{1-2x}{x}\le0\Rightarrow\left[{}\begin{matrix}x\ge\frac{1}{2}\\x< 0\end{matrix}\right.\)
c/ \(\left|2x+1\right|< 3x\)
- Với \(x< 0\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm
- Với \(x>0\Rightarrow2x+1>0\)
\(BPT\Leftrightarrow2x+1< 3x\Rightarrow x>1\)
d/ \(\sqrt{3x+1}\le x+1\)
ĐKXĐ: \(x\ge-\frac{1}{3}\)
DO 2 vế của BPT ko âm, bình phương 2 vế:
\(\left(x+1\right)^2\ge3x+1\)
\(\Leftrightarrow x^2-x\ge0\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\)
Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}-\frac{1}{3}\le x\le0̸\\x\ge1\end{matrix}\right.\)