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a: ĐKXĐ: (x+4)(x-1)<>0
hay \(x\notin\left\{-4;1\right\}\)
b: \(y-3=\dfrac{2x^2+6\sqrt{\left(x^2+1\right)\left(x-2\right)}+5-3x^2-9x+12}{x^2+3x-4}\)
\(=\dfrac{-x^2-9x+17+6\sqrt{\left(x^2+1\right)\left(x-2\right)}}{x^2+3x-4}< =0\)
=>y<=3
a) \(D=(0;+\infty)\backslash\left\{1\right\}\)
b) \(D=[2;+\infty)\)
\(y=2^{\sqrt{\left|x-3\right|-\left|8-x\right|}}+\sqrt{\frac{-\log_{0,5}\left(x-1\right)}{\sqrt{x^2-2x+8}}}\)
Điều kiện : \(\begin{cases}\left|x-3\right|-\left|8-x\right|\ge0\\\frac{-\log_{0,5}\left(x-1\right)}{\sqrt{x^2-2x+8}}\ge0\end{cases}\)
\(\Leftrightarrow\begin{cases}\left|x-3\right|\ge\left|8-x\right|\\x^2-2x-8>0\\\log_{0,5}\left(x-1\right)\le0\end{cases}\) \(\Leftrightarrow\begin{cases}\left(x-3\right)^2\ge\left(8-x\right)^2\\x^2-2x-8>0\\x-1\ge1\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge\frac{11}{2}\\x< -2;x>4\\x\ge2\end{cases}\)
\(\Leftrightarrow x\ge\frac{11}{2}\) là tập xác định của hàm số
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
Ta có
\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)
\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)
\(=54-2\sqrt{529}=8\)
\(\Rightarrow\) \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)
Xét tử số
\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)
\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23.2\sqrt{2}=46\sqrt{2}\)
Lại có \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)
\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)
\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)
ta bình phương mẫu số
\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)
\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)
Vậy mẫu \(=\sqrt{2}\)
Vậy \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\) thay vào ta đc A = 92880
a/ ĐKXĐ: \(\left\{{}\begin{matrix}2x+1\ge0\\3\left|x\right|^2+5\left|x\right|-2\ne0\\x-\left|x\right|\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\\left|x\right|\ne\frac{1}{3}\\x< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\frac{1}{2}\le x< 0\\x\ne-\frac{1}{3}\end{matrix}\right.\)
b/ Nếu \(x\in D\Rightarrow-x\in D\)
\(f\left(-x\right)=\frac{\left|-2017x-10\right|-\left|-2017x+10\right|}{x^6-8x^4+16x^2}\)
\(=\frac{\left|2017x+10\right|-\left|2017x-10\right|}{x^6-8x^4+16x^2}=-\frac{\left|2017x-10\right|-\left|2017x+10\right|}{x^6-8x^4+16x^2}=-f\left(x\right)\)
Hàm lẻ