\(^{x^6-\dfrac{5x^3}{3}+4\sqrt{x}-\dfrac{1}{x}+3}\)=?
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Những câu hỏi liên quan
25 tháng 9 2021
1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)
4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)
5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)
6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)
25 tháng 7 2023
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)
3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)
30 tháng 8 2021
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
MH
6 tháng 2 2022
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
TT
1 tháng 7 2018
b)\(x+\dfrac{4}{x+2}=3\) (ĐKXĐ: \(x\ne-2\))
\(\Leftrightarrow x+\dfrac{4}{x+2}-3=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)+4-3\left(x+2\right)}{x+2}=0\\ \Leftrightarrow\dfrac{x^2+2x+4-3x-6}{x+2}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{x+2}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x=\dfrac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4.1.\left(-2\right)}}{2.1}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{1+8}}{2}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{9}}{2}\\ \Leftrightarrow x=\dfrac{1\pm3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+3}{2}\\x=\dfrac{1-3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy...
28 tháng 8 2023
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 9 2021
a.
\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)
b.
\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)
c.
\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)
d.
\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)
e.
\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)
g.
\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)
\(y=x^6-\dfrac{5}{3}x^3+4\sqrt{x}-\dfrac{1}{x}+3\)
=>\(y'=6x^5-\dfrac{5}{3}\cdot3x^2+4\cdot\dfrac{1}{2\sqrt{x}}+\dfrac{1}{x^2}\)
=>\(y'=6x^5-5x^2+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x^2}\)
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