K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 7 2021

\(2sin^2\dfrac{x}{2}=cos5x+1\)

\(\Leftrightarrow-cos5x=1-2.sin^2\dfrac{x}{2}\)

\(\Leftrightarrow-cos5x=cosx\)

\(\Leftrightarrow cos\left(5x\right)=cos\left(\pi-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\pi-x+k2\pi\\5x=-\pi+x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\) (k nguyên)

Vậy..

8 tháng 4 2022

2.B (t/c của giới hạn)

6.B H/s ko x/đ với x = 0 -> Ko liên tục tại đ x = 0 

17.C

24. \(\lim\limits_{x\rightarrow\left(-1\right)^-}\dfrac{2x+1}{x+1}\)  . Thấy : \(\lim\limits_{x\rightarrow\left(-1\right)^-}2x+1=2.\left(-1\right)+1=-1\)

\(\lim\limits_{x\rightarrow\left(-1\right)^-}x+1=0\)  ; \(x\rightarrow\left(-1\right)^-\Rightarrow x+1< 0\).

Do đó : \(\lim\limits_{x\rightarrow\left(-1\right)^-}=+\infty\)  . Chọn B 

8 tháng 4 2022

33 . B 

Trên (SAB) ; Lấy H là TĐ của AB ; ta có : SH \(\perp AB\)  ( \(\Delta SAB\) đều ) ; HC \(\perp AB\) ( \(\Delta ABC\) đều ) 

Ta có : (SAB) \(\perp\left(ABC\right)\)  ; \(\left(SAB\right)\cap\left(ABC\right)=AB;SH\perp AB\)

\(\Rightarrow SH\perp\left(ABC\right)\)

\(SC\cap\left(ABC\right)=C\) . Suy ra : \(\left(SC;\left(ABC\right)\right)=\widehat{SCH}\)

Có : \(SH\perp HC\) => \(\Delta SHC\) vuông tại H 

G/s \(\Delta\)ABC đều có cạnh là a \(\Rightarrow AB=a\)

\(\Delta SAB\) đều => SA = SB = AB = a 

Tính được : \(SH=HC=\dfrac{\sqrt{3}}{2}a\)

\(\Delta SHC\) vuông tại H : \(tan\widehat{SCH}=\dfrac{SH}{HC}=1\)

\(\Rightarrow\widehat{SCH}=45^o\) => ... 

25 tháng 9 2021

\(2cos^2x-4sinxcosx=0\) 

\(\left\{{}\begin{matrix}cosx=0\\cosx-2sinx=0\end{matrix}\right.\)\(\left\{{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\cos\left(\alpha+x\right)=0vớicos\alpha=\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)

4 tháng 7 2021

ĐK: `x \ne kπ`

`cot(x-π/4)+cot(π/2-x)=0`

`<=>cot(x-π/4)=-cot(π/2-x)`

`<=>cot(x-π/4)=cot(x-π/2)`

`<=> x-π/4=x-π/2+kπ`

`<=>0x=-π/4+kπ` (VN)

Vậy PTVN.

1 tháng 8 2021

hahihihihi

17 tháng 9 2021

\(sin^2x+sin^22x=1\)

\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)

\(\Leftrightarrow-cos2x-2cos^22x+1=0\)

\(\Leftrightarrow\left(cos2x+1\right)\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

NV
23 tháng 3 2022

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

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

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

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

23 tháng 3 2022

em cảm ơn ạ

NV
9 tháng 3 2022

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

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

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

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

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

11 tháng 3 2022

em cảm ơn ạ

NV
25 tháng 7 2021

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.

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

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

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

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

NV
25 tháng 7 2021

3.

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

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

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

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

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

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

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

NV
10 tháng 7 2021

\(\Leftrightarrow2sin^3x+1-sin^2x-1=0\)

\(\Leftrightarrow sin^2x\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)