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Ta có : \(P=\frac{2a+3b+3c+1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c-1}{2017+c}\)
\(\Rightarrow P+3=\frac{2a+3b+3c+1}{2015+a}+1+\frac{3a+2b+3c}{2016+b}+1+\frac{3a+3b+2c-1}{2017+c}+1\)
\(=\frac{3a+3b+3c+2016}{2015+a}+\frac{3a+3b+3c+2016}{2016+b}+\frac{3a+3b+3c+2016}{2017+c}\)
\(=\left(3a+3b+3c+2016\right)\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)
\(=4.2016\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\) \(\left(a+b+c=2016\right)\)
\(=8064.\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)
Vì a ; b ; c dương , áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\), ta có :
\(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\ge\frac{9}{2015+2016+2017+a+b+c}=\frac{9}{8064}\)
\(\Rightarrow P+3\ge8064.\frac{9}{8064}=9\) \(\Rightarrow P\ge6\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2015+a=2016+b=2017+c\\a+b+c=2016\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1=c+2\\a+b+c=2016\end{matrix}\right.\)
\(\Leftrightarrow a=673;b=672;c=671\)
Vậy ...
Bài 1:
\(N=\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2017\)
\(=x^{2n}-2x^n+x^n-2-x^{2n}+x^n+2017\)
\(=2017\)
\(\Rightarrowđpcm\)
Bài 2:
\(A=-2\left(n+1\right)+n\left(2n-3\right)\)
\(=-2n^2-2n+2n^2-3n\)
\(=-5n⋮5\forall n\in Z\)
\(\Rightarrowđpcm\)
Bài 3:
\(A=x^8-2017x^7+2017x^6-2017x^5+...-2017x+2017\)
\(=x^8-2016x^7-x^7+2016x^6+x^6-2016x^5-x^5+2016x^4+...-2016x-x+2016+1\)
\(=x^7\left(x-2016\right)-x^6\left(x-2016\right)+x^5\left(x-2016\right)-x^4\left(x-2016\right)+...-\left(x-2016\right)+1\)
\(=\left(x^7-x^6+x^5-x^4+...-1\right)\left(x-2016\right)+1\)
Thay x = 2016
\(\Rightarrow A=1\)
Vậy A = 1 khi x = 2016
Câu 1:
a) 6x2 - 6xy
= 6x(x - y)
b) 9 + 2xy - x2 - y2
= -[(x2 - 2xy + y2) - 9]
= -[(x - y)2 - 32 ]
= -(x - y -3)(x - y + 3)
Câu 2:
a) 3x(x - 1) + (1 - x) = 0
3x2 - 3x + 1 - x = 0
3x(x - 1) - (x - 1) = 0
(x - 1)(3x - 1) = 0
=> x - 1 = 0 hoặc 3x - 1 = 0
TH1: x - 1 = 0
x = 1
TH2: 3x - 1 = 0
3x = 1
x = \(\frac{1}{3}\)
Vậy x ϵ {1; \(\frac{1}{3}\)}
b) x3 + 4x = 0
x (x2 + 4) = 0
=> x = 0 hoặc x2 + 4 = 0
=> x = 0 hoặc x2 = -4(Vô lí)
Vậy x = 0
c) Ko làm đc
Bài 1:
a) Đặt \(a=\dfrac{1}{229},b=\dfrac{1}{433}\), ta được
\(M=3a\left(2+b\right)-a\left(1-b\right)-4ab\)
\(M=6a+3ab-a+ab-4ab\)
\(M=5a\)
b) Ta có:
\(M=5a\)
\(M=\dfrac{5}{229}\)
Bài 2:
\(x=16\)
\(\Rightarrow x+1=17\left(1\right)\)
Thay (1) vào P, ta được:
\(P=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1+3\)
\(P=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1+3\)
\(P=4\)
Bài 3:
\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)
Vậy biểu thức không phụ thuộc vào x
Bài 4:
\(a\left(x-y\right)+b\left(y-x\right)\)
\(=a\left(x-y\right)-b\left(x-y\right)\)
\(=\left(x-y\right)\left(a-b\right)\)
Bài 5:
a) \(a.a^2.a^3.a^4.a^5a^6...a^{150}\)
\(=a^{1+2+3+4+5+6+...+150}\)
Đặt \(A=1+2+3+...+150\)
\(A=\dfrac{150-1+1}{2}\left(1+150\right)\)
\(A=75.151\)
\(A=2265\)
Vậy 1 + 2 + 3 +...+ 150 = 2265 (1)
Thay (1) vào ta được
\(a^{1+2+3+4+5+6+...+150}=a^{2265}\)
b) \(x^{2-k}.x^{1-k}.x^{2k-3}\)
\(=x^{2-k+1-k+2k-3}\)
\(=x^0\)
\(=1\)
Bài 6:
a) \(P=x\left(5x+15y\right)-5y\left(3x-2y\right)-5\left(y^2-2\right)\)
\(P=5x^2+15xy-15xy+10y^2-5y^2+10\)
\(P=5x^2+5y^2+10\)
b) \(P=0\)
\(\Rightarrow5x^2+5y^2+10=0\)
\(\Rightarrow5\left(x^2+y^2+2\right)=0\)
\(\Rightarrow x^2+y^2+2=0\)
\(\Rightarrow x^2+y^2=-2\)
Vì \(x^2\ge0\)
\(y^2\ge0\)
\(\Rightarrow x^2+y^2\ge0\)
Mà \(x^2+y^2=-2\)
=> Không tồn tại cặp số x và y để P = 0
\(P=10\)
\(\Rightarrow5x^2+5y^2+10=10\)
\(\Rightarrow5x^2+5y^2=0\)
\(\Rightarrow5\left(x^2+y^2\right)=0\)
\(\Rightarrow x^2+y^2=0\)
Vì \(x^2\ge0\) với mọi x
\(y^2\ge0\) với mọi y
\(\Rightarrow x^2+y^2\ge0\)
Mà \(x^2+y^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\ \Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\\ \Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\\ \Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\\ \Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\\ \Leftrightarrow\left(x+2005\right)=0\Leftrightarrow x=-2005\)
câu egf làm tương tự
-Mình nhận phần đại số nhé!!!
1.
a,\(5x\left(3x^2-4x+1\right)\)
\(=15x^3-20x^2+5x\)
b,\(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
2,
a,\(2x^2-4x\)
\(=2x\left(x-2\right)\)
b,\(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
c,\(x^2+x-6\)
\(=x^2+3x-2x-6\)
\(=\left(x^2-2x\right)+\left(3x-6\right)\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
3,
a,A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x+2\ne0\\x-2\ne0\\x^2-4\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)
Vậy...
b,\(A=\dfrac{1}{x+2}-\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}\)
\(=\dfrac{x-2}{x^2-4}-\dfrac{x+2}{x^2-4}+\dfrac{2x}{x^2-4}\)
\(=\dfrac{x-2-x-2-2x}{x^2-4}=\dfrac{-2x-4}{x^2-4}=\dfrac{-2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-2}{x-2}\)
c,\(A=\dfrac{-2}{x-2}\)
Thay \(x=\dfrac{1}{2}\) vào biểu thức,ta có:
\(A=\dfrac{-2}{\dfrac{1}{2}-2}=\dfrac{4}{3}\)
\(A=\frac{|x-2016|+2017}{|x-2016|+2018}\)
Ta thấy \(|x-2016|\ge0\forall x\)
\(\Rightarrow A=\frac{|x-2016|+2017}{|x-2016|+2018}\ge\frac{0+2017}{0+2018}\)
\(\Rightarrow A\ge\frac{2017}{2018}\)
\(\Rightarrow GTNN\)\(A=\frac{2017}{2018}\)