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Thay `x=2021` vào A: `A=2020.2021-2022 .2021^2 +2021^3=-2021`
a) A= (x-3)^2
thay x=203 vào 3, ta có
A=(203-3)^2=200^2=40000
\(x=2020\\ \Leftrightarrow x+1=2021\)
Thay vào biểu thức:
\(=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\\ =x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1=1\)
x + 3y = 10 <=> x = 10 - 3y thay vào D ta được:
D = (10 - 3y)2 + y2 = 100 - 60y + 9y2 + y2
D = 10y2 - 60y + 100 = 10(y2 - 6y + 10)
D = 10(y2 -2y3 + 9 + 1) = 10[(y - 3)2 + 1]
D = 10(y - 3)2 + 10 \(\ge\)10
Dấu "=" xảy ra khi: y - 3 = 0 <=> y = 3
=> x = 10 - 3y = 10 - 3.3= 1
Vậy gtnn D = 10 khi x = 1, y = 3
Lời giải:
Vì $x=9$ nên $x-9=0$
Ta có:
$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$
$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$
$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$
$=x-10=9-10=-1$
\(a,=\left(x+3\right)^3=\left(7+3\right)^3=10^3=1000\\ b,=\left(4-x\right)^3=\left(4-24\right)^3=\left(-20\right)^3=-8000\\ c,=\left(x-1\right)^3=\left(11-1\right)^3=10^3=1000\)
\(A=\frac{3x^2+8x+6}{x^2+2x+1}\) \(\left(x\ne\pm1\right)\)
\(A=\frac{\left(3x^2+6x+3\right)+\left(2x+3\right)}{\left(x+1\right)^2}\)
\(A=\frac{3\left(x+1\right)^2+2x+3}{\left(x+1\right)^2}\)
\(A=3+\frac{2x+3}{\left(x+1\right)^2}\)
Vì\(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow3+\frac{2x+3}{\left(x+1\right)^2}\ge3\Leftrightarrow A\ge3\)
Dấu "="xảy ra khi \(2x+3=0\Rightarrow x=\frac{-3}{2}\)
Gọi k là một giá trị của A ta có:
\(\frac{\left(3x^2-8x+6\right)}{\left(x^2+2x+1\right)}=k\)
\(\Leftrightarrow3x^2-8x+6=k\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(3-k\right)x^2-\left(8-2k\right)x+6-k=0\)(*)
Ta cần tìm k để PT (*) có nghiệm
Xét: \(\Delta=\left(8-2k\right)^2-4\left(3-k\right)\left(6-k\right)=64-32k+4k^2-4\left(18-9k+k^2\right)=4k-8\)
Để PT (*) có nghiệm thì: \(\Delta\ge0\Leftrightarrow4k-8\ge0\Leftrightarrow k\ge2\)
Dấu "=" xảy ra khi: \(-\left(8-2.2\right)x+6-2=0\Leftrightarrow-4x+4=0\Rightarrow x=1\)
Vậy: \(B\ge2\)suy ra: B = 2 khi x = 1
a) Ta có: \(A=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)
\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{2x^2+1}{x^2+1}\cdot\dfrac{\left(x-1\right)\left(x^2+1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2+1}{x-1}\)
b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)
\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
c) Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)
\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)
\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)
hay x<1
\(x=2021\Leftrightarrow x+1=2022\\ \Leftrightarrow P=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x\\ P=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x\\ P=0\)
\(P=x^5-2022x^4+2022x^3-2022x^2+2022x-2021=x^4\left(x-2021\right)-x^3\left(x-2021\right)+x^2\left(x-2021\right)-x\left(x-2021\right)+\left(x-2021\right)\)
\(=\left(x-2021\right)\left(x^4-x^3+x^2-x+1\right)\)
\(=\left(2021-2021\right)\left(x^4-x^3+x^2-x+1\right)=0\)