Cho A = 22 - 3x + 5. Tính S tại |x| = \(\dfrac{1}{2}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)
2: \(=\dfrac{3x+1-2x^2-2x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
a: \(\dfrac{5x}{x^2+x-6}=\dfrac{5x}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{4x+12+x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{4}{x-2}+\dfrac{x-12}{x^2+x-6}\)
=>\(A=\dfrac{4}{x-2};B=\dfrac{x-12}{x^2+x-6}\)
b: \(\dfrac{5x+31}{x^2-3x-10}=\dfrac{5x+31}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x-15+2x+46}{\left(x-5\right)\left(x+2\right)}=\dfrac{3}{x+2}+\dfrac{2x+46}{\left(x-5\right)\left(x+2\right)}\)
=>\(A=\dfrac{3}{x+2};B=\dfrac{2x+46}{\left(x-5\right)\left(x+2\right)}\)
c: \(\dfrac{3x+5}{\left(x-1\right)^2}=\dfrac{3x-3+8}{\left(x-1\right)^2}=\dfrac{3}{x-1}+\dfrac{8}{\left(x-1\right)^2}\)
=>\(A=\dfrac{3}{x-1};B=\dfrac{8}{\left(x-1\right)^2}\)
\(a.3x-5y+1=3.\dfrac{1}{3}-5.\left(-\dfrac{1}{5}\right)+1=1+1+1=3\)
b.x=1
\(\Rightarrow3.1^2-2.1-5=-4\)
x=-1
\(\Rightarrow3.\left(-1\right)^2-2.\left(-1\right)-5=3+2-5=0\)
Đặt `A=(1-3x)/(2x)+(3x-2)/(2x-1)+(3x-2)/(2x-4x^2)`
`=(2x(3x-2))/(2x(2x-1))-((3x-1)(2x-1))/(2x(2x-1))-(3x-2)/(2x(2x-1))`
`=(6x^2-4x-6x^2+5x-1-3x+2)/(2x(2x-1))`
`=(-2x+1)/(2x(2x-1))`
`=-1/(2x)`
`2x=1/(483)`
`=>A=-1/(1/483)=-483`
a. Làm gọn 1 chút xíu:
\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)
\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)
b.
\(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)
c.
\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)
S=5/4(x-1)+5/x-1+9/4(y-1)+9/(y-1)+7/4(x+y)+7/2
=5/4(x-1)+5/(x-1)+9/4(y-1)+9/y-1+14
=>S>=2*5/2+2*9/2+14=28
Dấu = xảy ra khi x=y=3
Lời giải:
$x+\frac{1}{x}=4\Rightarrow x^2+1=4x$.
$S=\frac{x^6+3x^3+1}{x^2(x^2+1)}=\frac{(x^6+1)+3x^3}{4x^3}$
$=\frac{1}{4}(x^3+\frac{1}{x^3})+\frac{3}{4}$
$=\frac{1}{4}[(x+\frac{1}{x})^3-3x.\frac{1}{x}(x+\frac{1}{x})]+\frac{3}{4}$
$=\frac{1}{4}(4^3-3.4)+\frac{3}{4}=\frac{55}{4}$
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)