a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)

A. √3+1/2         B. √3−1/2          C. 1−√3/2             D. 0

B

\(f'\left(x\right)-f\left(x\right)=2cosx\)

\(\Leftrightarrow e^{-x}.f'\left(x\right)-e^{-x}.f\left(x\right)=2e^{-x}cosx\)

\(\Rightarrow\left[e^{-x}.f\left(x\right)\right]'=2e^{-x}.cosx\)

Lấy nguyên hàm 2 vế:

\(\Rightarrow e^{-x}.f\left(x\right)=\int2e^{-x}cosxdx=e^{-x}\left(sinx-cosx\right)+C\)

Thay \(x=\dfrac{\pi}{2}\Rightarrow e^{-\dfrac{\pi}{2}}.1=e^{-\dfrac{\pi}{2}}+C\Rightarrow C=0\)

\(\Rightarrow f\left(x\right)=sinx-cosx\)

\(\Rightarrow f\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}-1}{2}\)

a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)

\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)

b: \(y=\left(3x+1\right)^{\Omega}\)

=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)

=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)

c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)

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

\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)

d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)

\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)

\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)

\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)

e: \(y=3^{x^2}\)

=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)

f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)

h: \(y=\left(x+1\right)\cdot e^{cosx}\)

=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)

=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)

\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)

a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)

b) \(y=\left(3x+1\right)^{\pi}\)

\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)

c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)

d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)

e) \(y=3^{x^2}\)

\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)

f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)

Các bài còn lại bạn tự làm nhé!

Làm xuôi thì đơn giản, tính \(F'\left(x\right)\) là xong (chịu khó biến đổi)

Làm ngược thì nhìn biểu thức hơi thiếu thân thiện

\(\int\dfrac{2\sqrt{2}\left(x^2-1\right)}{x^4+1}dx=\int\dfrac{2\sqrt{2}\left(x^2-1\right)}{\left(x^2-x\sqrt{2}+1\right)\left(x^2+x\sqrt{2}+1\right)}dx\)

Phân tách hệ số bất định:

\(\dfrac{2\sqrt{2}\left(x^2-1\right)}{\left(x^2-x\sqrt{2}+1\right)\left(x^2+x\sqrt{2}+1\right)}=\dfrac{a\left(2x-\sqrt{2}\right)}{x^2-x\sqrt{2}+1}+\dfrac{b\left(2x+\sqrt{2}\right)}{x^2+x\sqrt{2}+1}\)

Quan tâm tử số: \(a\left(2x-\sqrt{2}\right)\left(x^2+x\sqrt{2}+1\right)+b\left(2x+\sqrt{2}\right)\left(x^2-x\sqrt{2}+1\right)\)

\(=2\left(a+b\right)x^3+\sqrt{2}\left(a-b\right)x^2+\sqrt{2}\left(b-a\right)\)

Đồng nhất 2 tử số: \(\left\{{}\begin{matrix}a+b=0\\a-b=2\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-1\end{matrix}\right.\)

Do đó:

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

Đặt \(tan\left(x+\dfrac{\pi}{3}\right)=t\)

\(\Rightarrow t^2+\left(\sqrt{3}-1\right)t-\sqrt{3}=0\)

\(\Leftrightarrow t\left(t-1\right)+\sqrt{3}\left(t-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tan\left(x+\dfrac{\pi}{3}\right)=1\\tan\left(x+\dfrac{\pi}{3}\right)=-\sqrt{3}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=-\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\)

Câu 2:

\(\Leftrightarrow\dfrac{\left(n+2\right)!}{2!\cdot n!}-4\cdot\dfrac{\left(n+1\right)!}{n!\cdot1!}=2\left(n+1\right)\)

\(\Leftrightarrow\dfrac{\left(n+1\right)\left(n+2\right)}{2}-4\cdot\dfrac{n+1}{1}=2\left(n+1\right)\)

\(\Leftrightarrow\left(n+1\right)\left(n+2\right)-8\left(n+1\right)=4\left(n+1\right)\)

=>(n+1)(n+2-8-4)=0

=>n=-1(loại) hoặc n=10

=>\(A=\left(\dfrac{1}{x^4}+x^7\right)^{10}\)

SHTQ là: \(C^k_{10}\cdot\left(\dfrac{1}{x^4}\right)^{10-k}\cdot x^{7k}=C^k_{10}\cdot1\cdot x^{11k-40}\)

Số hạng chứa x^26 tương ứng với 11k-40=26

=>k=6

=>Số hạng cần tìm là: \(210x^{26}\)

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

\(=\dfrac{1}{0}=+\infty\)

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

\(f\left(1\right)=\sqrt{2}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)\ne\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne f\left(x\right)\)=> ham gian doan tai x=1

a: =>2sin(x+pi/3)=-1

=>sin(x+pi/3)=-1/2

=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi

=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi

b: =>2sin(x-30 độ)=-1

=>sin(x-30 độ)=-1/2

=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ

=>x=k*360 độ hoặc x=240 độ+k*360 độ

c: =>2sin(x-pi/6)=-căn 3

=>sin(x-pi/6)=-căn 3/2

=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi

=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi

d: =>2sin(x+10 độ)=-căn 3

=>sin(x+10 độ)=-căn 3/2

=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ

=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ

e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)

=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)

=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ

=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ

f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)

=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi

=>x=pi/12+k2pi hoặc x=19/12pi+k2pi

g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

h) \(1+sin\left(x-30^o\right)=0\)

\(\Leftrightarrow sin\left(x-30^o\right)=-1\)

\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow x=-60^0+k360^o\)

a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)

\(\Rightarrow3x+60=30+2k180\)

\(\Rightarrow3x=2k180-30\)

\(\Leftrightarrow x=120k-10\)

Vậy ...

b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)

Vậy ...

c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)

\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Vậy ...

d, Ta có : \(\cot\left(2x+\pi\right)=-1\)

\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)

\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)

Vậy ...

a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{3}\right)=sin\dfrac{\pi}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)

Vậy...

d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)

\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)

\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)

Vậy...

a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)

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

b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)

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

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)

c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)

\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)

=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{\Omega}{3}+k\Omega\)

d: ĐKXĐ: \(2x< >k\Omega\)

=>\(x< >\dfrac{k\Omega}{2}\)

\(cot2x=-\dfrac{\sqrt{3}}{3}\)

=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)

=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)

=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)