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1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)
2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\) \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)
3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)
\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)
b.
ĐKXĐ: \(x\ge-1\)
\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)
\(\Leftrightarrow...\)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+2a^2=-b^2+b+3ab\)
\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)
\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)
\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)
\(a,3^{x-1}=27\\ \Leftrightarrow3^{x-1}=3^3\\ \Leftrightarrow x-1=3\\ \Leftrightarrow x=4\\ b,100^{2x^2-3}=0,1^{2x^2-18}\\ \Leftrightarrow10^{4x^2-6}=10^{-2x^2+18}\\ \Leftrightarrow4x^2-6=-2x^2+18\\ \Leftrightarrow6x^2=24\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)
\(c,\sqrt{3}e^{3x}=1\\ \Leftrightarrow e^{3x}=\dfrac{1}{\sqrt{3}}\\ \Leftrightarrow3x=ln\left(\dfrac{1}{\sqrt{3}}\right)\\ \Leftrightarrow x=\dfrac{1}{3}ln\left(\dfrac{1}{\sqrt{3}}\right)\)
\(d,5^x=3^{2x-1}\\ \Leftrightarrow2x-1=log_35^x\\ \Leftrightarrow2x-1-xlog_35=0\\ \Leftrightarrow x\left(2-log_35\right)=1\\ \Leftrightarrow x=\dfrac{1}{2-log_35}\)
a: \(2^{x^2-2x+1}=1\)
=>\(2^{\left(x-1\right)^2}=2^0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
b: \(7^{x^2+7x}=5764801\)
=>\(7^{x^2+7x}=7^8\)
=>\(x^2+7x=8\)
=>\(x^2+7x-8=0\)
=>(x+8)(x-1)=0
=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
c: \(6^{x^2+12x}=6^{7x}\)
=>\(x^2+12x=7x\)
=>\(x^2+5x=0\)
=>x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)
=>\(3^{-x+1}=3^{2x-5}\)
=>-x+1=2x-5
=>-x-2x=-5-1
=>-3x=-6
=>x=2
e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)
=>\(5^{-3x-5}=5^{2x+1}\)
=>-3x-5=2x+1
=>-5x=6
=>\(x=-\dfrac{6}{5}\)
a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)
b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)
c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)