Tìm GTLN của A= \(4x^2+4x+2016\) B=\(\dfrac{-7}{x^2+6x+2012}\) C= \(\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
Bài 2 . Thực hiện phép tính
a)\(6x^3\)\(\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)\)\(-2x^5\)\(-x^3\)
b)\(\left(x-3\right)\left(x^2+3x-2\right)\)
c)\(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
Chứng minh rằng \(f'\left(x\right)=0;\forall x\in R\) nếu :
a) \(f\left(x\right)=3\left(\sin^4x+\cos^4x\right)-2\left(\sin^6x+\cos^6x\right)\)
b) \(f\left(x\right)=\cos^6x+2\sin^4x.\cos^2x+3\sin^2x\cos^4x+\sin^4x\)
c) \(f\left(x\right)=\cos\left(x-\dfrac{\pi}{3}\right)\cos\left(x+\dfrac{\pi}{4}\right)+\cos\left(x+\dfrac{\pi}{6}\right)\cos\left(x+\dfrac{3\pi}{4}\right)\)
d) \(f\left(x\right)=\cos^2x+\cos^2\left(\dfrac{2\pi}{3}+x\right)+\cos^2\left(\dfrac{2\pi}{3}-x\right)\)
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
tìm GTLN
a)\(A=x^2+5y^2+2xy-4x-8y+2015\)
b)\(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
c)\(C=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)
d)\(D=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)
\(B=2x^2+2x+2012^2+2013^2\)
\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)
\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)
\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)
Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)
Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
a: \(=\dfrac{x-z}{2}\)
b: \(=\dfrac{3x}{4y^3}\)
a) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{3}-\left(x-1\right)\)
b) \(x^2-6x-2+\dfrac{14}{x^2-6x+7}=0\)
c) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
d) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}=\dfrac{6}{x^2-9}\)
e) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
Chứng minh các biểu thức sau không phụ thuộc vào x:
1, \(A=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
2, \(B=cos^6x+2sin^4x.cos^2x+3sin^2x.cos^4x+sin^4x\)
3, \(C=cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
4, \(D=cos^2x+cos^2\left(x+\dfrac{2\pi}{3}\right)+cos^2\left(\dfrac{2\pi}{3}-x\right)\)
5, \(E=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)-\left(sin^8x+cos^8x\right)\)
6, \(F=cos\left(\pi-x\right)+sin\left(\dfrac{-3\pi}{2}+x\right)-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\dfrac{3\pi}{2}-x\right)\)
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
Tìm GTNN của các biểu thức sau:
A = \(\left(x^2-x\right)\left(x^2+3x+2\right)\)
B = \(x^4+\left(x-2\right)^2+6x^2\left(x-2\right)^2\)
C = \(\frac{2x^2-6x+5}{\left(x-1\right)^2}\)
D = \(4x^2+4x-6\left|2x+1\right|+6\)
Tìm cả GTLN và GTNN
A = \(\frac{3-4x}{x^2+1}\)
Em làm bài 2 nha!
\(A=\frac{3-4x}{x^2+1}\Leftrightarrow Ax^2+4x+A-3=0\) (1)
+)\(A=0\Rightarrow x=\frac{3}{4}\)
+) A khác 0 thì (1) là pt bậc 2.
\(\Delta'=\left(2\right)^2-A\left(A-3\right)\ge0\Leftrightarrow4-A^2+3A\ge0\Leftrightarrow-1\le A\le4\)
Vậy...
Bài 1: (bài nào nghĩ ra thì em làm trước)
C = \(\frac{2x^2-6x+5}{\left(x-1\right)^2}\). Đặt x - 1 = y >0 thì x = y + 1 >1
Khi đó \(C=\frac{2\left(y+1\right)^2-6\left(y+1\right)+5}{y^2}=\frac{2y^2-2y+1}{y^2}\)
\(=\frac{1}{y^2}-\frac{2}{y}+2\). đặt \(\frac{1}{y}=t>0\). \(C=t^2-2t+2=\left(t-1\right)^2+1\ge1\)
Đẳng thức xảy ra khi t = 1 suy ra y = 1 suy ra x = 2
Vậy Min C = 1 khi x = 2
Tìm GTLN của: A=x/(x+10)^2 \(B=\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\)
1) Cho P = \(\left(\dfrac{4x-x^3}{1-4x^2}-x\right):\left(\dfrac{4x^2-x^4}{1-x^2}+1\right)\)
a) rút gọn b) tìm x để P > 0
2) Cho Q = \(\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{x^3+27}+\dfrac{1}{x+3}\right):\dfrac{x^2-1}{x+3}\)
a) rút gọn b) tìm GTLN
3) Cho A = \(\dfrac{1}{\left(x-y\right)^3}\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)+\dfrac{3}{\left(x-y\right)^4}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{6}{\left(x-y\right)^5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
chứng minh A là lập phương một số hữu tỉ