5. 

ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)

Pt tương đương:

\(\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow x=30^0+k180^0\)

6.

\(\Leftrightarrow2\sqrt{2}sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

1.

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

2.

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x-2}+1}{\sqrt[]{x+3}-2}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{x-2}+1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)\left(\sqrt[]{x+3}+2\right)}{\left(\sqrt[]{x+3}-2\right)\left(\sqrt[]{x+3}+2\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\sqrt[]{x+3}+2\right)}{\left(x-1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x+3}+2}{\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1}\)

\(=\dfrac{\sqrt[]{1+3}+2}{\sqrt[3]{\left(1-2\right)^2}-\sqrt[3]{1-2}+1}=\dfrac{4}{3}\)

em cảm ơn ạ

\(\lim\dfrac{3^n+2.6^n}{6^{n-1}+5.4^n}=\lim\dfrac{6^n\left[\left(\dfrac{3}{6}\right)^n+2\right]}{6^n\left[\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n\right]}=\lim\dfrac{\left(\dfrac{3}{6}\right)^n+2}{\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n}=\dfrac{0+2}{\dfrac{1}{6}+0}=12\)

\(\lim\left(\sqrt{n^2+9}-n\right)=\lim\dfrac{\left(\sqrt{n^2+9}-n\right)\left(\sqrt{n^2+9}+n\right)}{\sqrt{n^2+9}+n}=\lim\dfrac{9}{\sqrt{n^2+9}+n}\)

\(=\lim\dfrac{n\left(\dfrac{9}{n}\right)}{n\left(\sqrt{1+\dfrac{9}{n^2}}+1\right)}=\lim\dfrac{\dfrac{9}{n}}{\sqrt{1+\dfrac{9}{n^2}}+1}=\dfrac{0}{1+1}=0\)

\(\lim\dfrac{\sqrt{15+9n^2}-3}{5-n}=\lim\dfrac{n\sqrt{\dfrac{15}{n^2}+9}-3}{5-n}=\lim\dfrac{n\left(\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}\right)}{n\left(\dfrac{5}{n}-1\right)}\)

\(=\lim\dfrac{\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}}{\dfrac{5}{n}-1}=\dfrac{\sqrt{9}-0}{0-1}=-3\)

em cảm ơn ạ

7.

\(y'=3x^2+8x-1\)

\(\Rightarrow y'\left(2\right)=3.2^2+8.2-1=27\)

a: -1<=sin x<=1

=>-3<=3sin x<=3

=>-3+1<=3sin x+1<=3+1

=>-2<=y<=4

y min=-2 khi sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

y max=4 khi sinx=1

=>\(x=\frac{\pi}{2}+k2\pi\)

b: -1<=sin x<=1

=>1>=-sin x>=-1

=>1+4>=-sin x+4>=-1+4

=>5>=y>=3

y max=5 khi sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

y min=3 khi sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

c: \(0\le\sqrt{\sin x}\le1\)

=>\(0\le2\cdot\sqrt{\sin x}\le2\)

=>\(0+1\le2\cdot\sqrt{\sin x}+1\le2+1\)

=>1<=y<=3

y min=1 khi sin x=0

=>\(x=k\pi\)

y max=1 khi sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

d: \(-1\le cosx\le1\)

=>\(-1\cdot5\le5\cdot cosx\le1\cdot5\)

=>-5<=5cosx<=5

=>-5+1<=5*cosx+1<=5+1

=>-4<=y<=6

y min=-4 khi cosx=-1

=>\(x=\pi+k2\pi\)

y max=6 khi cosx=1

=>\(x=k2\pi\)

e: -1<=cosx<=1

=>2>=-2cosx>=-2

=>2+3>=-2*cosx+3>=-2+3

=>5>=y>=1

y max =5 khi cosx=-1

=>\(x=\pi+k2\pi\)

y min=1 khi cosx=1

=>\(x=k2\pi\)

f: \(0\le\sqrt{cosx}\le1\)

=>\(0\le3\cdot\sqrt{cosx}\le3\cdot1=3\)

=>\(0+4\le3\cdot\sqrt{cosx}+4\le3+4\)

=>4<=y<=7

y min=4 khi cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

y max=7 khi cosx=1

=>\(x=k2\pi\)

g: \(0\le\sin^2x\le1\)

=>\(0\le4\cdot\sin^2x\le4\)

=>\(0-1\le4\cdot\sin^2x-1\le4-1\)

=>-1<=y<=3

y min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

y max=3 khi \(\sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

h: \(0\le cos^2x\le1\)

=>\(0\le9\cdot cos^2x\le9\)

=>\(0-1\le9\cdot cos^2x-1\le9-1\)

=>-1<=y<=8

y min=-1 khi \(cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

y max=8 khi \(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

Chọn B