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NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2020
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2020
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
7 tháng 11 2018
1) \(y=\dfrac{2x^2+1}{x^3-5x+4}\)
ĐK \(x^3-5x+4\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\dfrac{\sqrt{17}-1}{2}\\x\ne\dfrac{-\sqrt{17}-1}{2}\end{matrix}\right.\)
TXĐ \(D=R\backslash\left\{1;\dfrac{\sqrt{17}-1}{2};\dfrac{-\sqrt{17}-1}{2}\right\}\)
2) \(y=\dfrac{\sqrt{x-2}}{\left(x-3\right)^3-1}\)
ĐK \(\left\{{}\begin{matrix}x-2\ge0\\x-3\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne4\end{matrix}\right.\)
TXĐ \(D=[2;+\infty)\backslash\left\{4\right\}\)
3) \(y=\sqrt{x-2}-\dfrac{2}{\sqrt[3]{x-1}}\)
ĐK\(\left\{{}\begin{matrix}x+2\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ne1\end{matrix}\right.\)
TXĐ \(D=[-2;+\infty)\backslash\left\{1\right\}\)
4) \(y=\dfrac{x^2+2}{\sqrt{\left(x+3\right)^2}}=\dfrac{x^2+2}{\left|x-3\right|}\)
ĐK \(x-3\ne0\Leftrightarrow x\ne3\)
TXĐ \(D=R\backslash\left\{3\right\}\)
5) \(y=\dfrac{\sqrt{x^2-2}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
ĐK \(\left\{{}\begin{matrix}x^2-2\ge0\\x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in(-\infty;-\sqrt{2}]\cap[\sqrt{2};+\infty)\\x>0\\x\ne9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge\sqrt{2}\\x\ne9\end{matrix}\right.\)
TXĐ \(D=[\sqrt{2};+\infty)\backslash\left\{9\right\}\)
6) \(y=\sqrt{1-\sqrt{1+x}}\)
ĐK \(\left\{{}\begin{matrix}x+1\ge0\\1-\sqrt{1+x}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge\sqrt{1+x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge1+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le0\end{matrix}\right.\)
TXĐ \(D=\left[0;-1\right]\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2020
\(x-\frac{11x^2-5x+6}{x^2+5x+6}>0\)
\(\Leftrightarrow\frac{x^3-6x^2+11x-6}{x^2+5x+6}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)}{\left(x+2\right)\left(x+3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\1< x< 2\\-3< x< -2\end{matrix}\right.\)
b/ \(\frac{2-x}{x^3+x^2}-\frac{1-2x}{x^3-3x^2}>0\)
\(\Leftrightarrow\frac{\left(2-x\right)\left(x+1\right)-\left(1-2x\right)\left(x-3\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-5\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>5\\1< x< 3\end{matrix}\right.\)
c/ \(\left|x^2-x-1\right|\le x-1\)
Với \(x< 1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm
Với \(x\ge1\) hai vế ko âm, bình phương:
\(\left(x^2-x-1\right)^2\le\left(x-1\right)^2\)
\(\Leftrightarrow\left(x^2-x-1\right)^2-\left(x-1\right)^2\le0\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2\right)\le0\) \(\Rightarrow\sqrt{2}\le x\le2\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
(x + 6)(x + 3)(x + 9)(x + 2) = 5x2
<=> (x2 + 9x + 18).(x2 + 11x + 18) = 5x2
<=> (x2 + 10x + 18 - x)(x2 + 10x + 18 + x) = 5x2
<=> (x2 + 10x + 18)2 - x2 = 5x2
<=> (x2 + 10x + 18)2 = 6x2
<=> \(\left[{}\begin{matrix}x^2+10x+18=\sqrt{6}x\\x^2+10x+18=-\sqrt{6}x\end{matrix}\right.\)
Với \(x^2+10x+18=\sqrt{6}x\Leftrightarrow x^2+\left(10-\sqrt{6}\right)x+18=0\)
\(\Delta=\left(10-\sqrt{6}\right)^2-72=34-20\sqrt{6}< 0\)
=> Phương trình vô nghiệm
Với \(x^2+10x+18=-\sqrt{6}x\Leftrightarrow x^2+\left(10+\sqrt{6}\right)x+18=0\)
\(\Delta=\left(10+\sqrt{6}\right)^2-72=34+20\sqrt{6}\) > 0
Phương trình có 2 nghiệm \(x=\dfrac{-10-\sqrt{6}\pm\sqrt{34+20\sqrt{6}}}{2}\)
\(\left(x+6\right)\left(x+3\right)\left(x+9\right)\left(x+2\right)=5x^2\)
\(\Leftrightarrow\left(x^2+3x+6x+18\right)\left(x^2+2x+9x+18\right)=5x^2\)
\(\Leftrightarrow\left(x^2+9x+18\right)\left(x^2+11x+18\right)=5x^2\)
\(\Leftrightarrow x^4+11x^3+18x^2+9x^3+99x^2+162x+18x^2+198x+324=5x^2\)
\(\Leftrightarrow x^4+20x^3+135x^2+360x+324=5x^2\)
\(\Leftrightarrow x^4+20x^3+130x^2+360x+324=0\)
\(\Leftrightarrow x\in\varnothing\)