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\(A=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)\left(5x+4\right)^2\)
\(=\left(5x-1\right)-2\left(5x-1\right)\left(5x+4\right)^3\)
\(=\left(5x-1\right)\left(1-2\left(5x+4\right)^3\right)\)
\(=\left(5x-1\right)\left(1-2\left(125x^3+300x^2+240x+64\right)\right)\)
\(=\left(5x-1\right)\left(1-250x^3-600x^2-480x-128\right)\)
\(=5x-1250x^4-3000x^3-2400x^2-640x-1+250x^3+600x^2+480x+128\)
\(=-1250x^4-2750x^3-1800x^2-110x+127\)
(Số hơi to)
\(B=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(B=\left(x-y\right)^3+\left(y+x\right)^3-\left(x-y\right)^3-3xy\left(x+y\right)\)
\(B=\left(y+x\right)^3-3xy\left(x+y\right)\)
\(B=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
\(B=\left(x+y\right)\left[x^2+2xy+y^2-3xy\right]\)
\(B=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(A=-\left(x^2-2x+1\right)-2\)
\(A=-\left(x-1\right)^2-2\)
Vì \(-\left(x-1\right)^2\le0;\forall x\)
\(\Rightarrow-\left(x-1\right)^2-2\le0-2;\forall x\)
Hay \(A\le-2;\forall x\)
Dấu "=" xảy ra\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Vậy MAX A=-2 \(\Leftrightarrow x=1\)
\(C=-2x^2+2xy-y^2+2x+4\)
\(C=-x^2+2xy-y^2-x^2+2x-1+5\)
\(C=-\left(x^2-2xy+y^2\right)-\left(x^2-2x+1\right)+5\)
\(C=-\left(x-y\right)^2-\left(x-1\right)^2+5\le5\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\x=1\end{cases}}\Leftrightarrow x=y=1\)
Vậy C max = 5 tại x = y = 1
Đăng từng bài thoy nha pn!!!
Bài 1:
Có : 2009 = 2008 + 1 = x + 1
Thay 2009 = x + 1 vào biểu thức trên,ta có :
x\(^5\)- 2009x\(^4\)+ 2009x\(^3\)- 2009x\(^2\)+ 2009x - 2010
= x\(^5\)- (x + 1)x\(^4\)+ (x + 1)x\(^3\)- (x +1)x\(^2\)+ (x + 1) x - (x + 1 + 1)
= x\(^5\)- x\(^5\)- x\(^4\)+ x\(^4\)- x\(^3\)+ x\(^3\)- x\(^2\)+ x\(^2\)+ x - x -1 - 1
= -2
\(B=x^2-x+2=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{7}{4}=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
Vậy \(B_{min}=\frac{7}{4}\)\(\Leftrightarrow x=\frac{1}{2}\)
\(A=2x^2-3x+6=2\left(x^2-2.\frac{3}{4}x+\frac{9}{16}+\frac{39}{16}\right)\)
\(=2\left[\left(x-\frac{3}{4}\right)^2+\frac{39}{16}\right]\ge\frac{39}{8}\)
Vậy \(A_{min}=\frac{39}{8}\Leftrightarrow x=\frac{3}{4}\)
a) \(A=-|x-2|\le0;\forall x\)
\(\Rightarrow-|x-2|+2019\le0+2019;\forall x\)
Hay \(A\le2019;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow|x-2|=0\)
\(\Leftrightarrow x=2\)
Vậy \(A_{max}=2019\Leftrightarrow x=2\)
b) \(B=-2x^2+5x+3\)
\(=-2\left(x^2-\frac{5}{2}x-\frac{3}{2}\right)\)
\(=-2\left(x^2-2.x.\frac{5}{4}+\frac{25}{16}-\frac{25}{16}-\frac{3}{2}\right)\)
\(=-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\)
Vì \(-2\left(x-\frac{5}{4}\right)^2\le0;\forall x\)
\(\Rightarrow-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\le0+\frac{49}{8};\forall x\)
Hay \(B\le\frac{49}{8};\forall x\)
Dấu "="xảy ra \(\Leftrightarrow\left(x-\frac{5}{4}\right)^2=0\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy \(B_{max}=\frac{49}{8}\Leftrightarrow x=\frac{5}{4}\)
c) \(-x^2-y^2+2x+8y+2028\)
\(=-\left(x^2+y^2-2x-8y-2028\right)\)
\(=-\left[\left(x^2-2x+1\right)+\left(y^2-8y+16\right)-2045\right]\)
\(=-\left(x-1\right)^2-\left(y-4\right)^2+2045\)
Vì \(\hept{\begin{cases}-\left(x-1\right)^2\le0;\forall x,y\\-\left(y-4\right)^2\le0;\forall x,y\end{cases}}\)
\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2\le0;\forall x,y\)
\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2+2045\le0+2045;\forall x,y\)
Hay \(C\le2045;\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}-\left(x-1\right)^2=0\\-\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}}\)
Vậy \(C_{max}=2045\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}\)
Ta có: P = 2(x6 + y6) - 3(x4 + y4)
P = 2(x2 + y2)(x4 - x2y2 + y4) - 3x4 - 3y4
P = 2.1.(x4 - x2y2 + y4) - 3x4 - 3y4
P = 2x4 - 2x2y2 + 2y4 - 3x4 - 3y4
P = (2x4 - 3x4) - 2x2y2 + (2y4 - 3y4)
P = -x4 - 2x2y2 - y4
P = -(x4 + 2x2y2 + y4)
P = -(x2 + y2)2
P = -12 = -1
=> Biểu thức P ko phụ thuộc vào x với x2 + y2 = 1