Question

[Phan Thị Hương Ly]

chứng minh biểu thức luôn dương

a) A=16x²+8x+3

b) B=y²-5y+8

c) C=2x²-2x+2

d) D=9x²-6x+25y²+10y+4

tìm Min hoặc Max của các biểu thức sau

a) M=x²+6x-1

b) N=10y-5y²-3

Giupa mình với nha

Answer 1

[Trần Quốc Lộc]

\(A=16x^2+8x+3\\ A=16x^2+8x+1+2\\ A=\left(16x^2+8x+1\right)+2\\ A=\left(4x+1\right)^2+2\\ Do\left(4x+1\right)^2\ge0\forall x\\ \Rightarrow A=\left(4x+1\right)^2+2\ge2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(4x+1\right)^2=0\\ \Leftrightarrow4x+1=0\\ \Leftrightarrow4x=-1\\ \Leftrightarrow x=-\dfrac{1}{4}\\ \text{Vậy }A_{\left(Min\right)}=2\text{ khi }x=-\dfrac{1}{4}\\ \)

\(B=y^2-5y+8\\ B=y^2-5y+\dfrac{25}{4}+\dfrac{7}{4}\\ B=\left(y^2-5y+\dfrac{25}{4}\right)+\dfrac{7}{4}\\ B=\left[y^2-2\cdot y\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{7}{4}\\ B=\text{ }\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\\ Do\text{ }\left(y-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow B=\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{5}{2}=0\\ \Leftrightarrow y=\dfrac{5}{2}\\ \text{Vậy }B_{\left(Min\right)}=\dfrac{7}{4}\text{ }khi\text{ }y=\dfrac{5}{2}\)

\(C=2x^2-2x+2\\ C=2x^2-2x+\dfrac{1}{2}+\dfrac{3}{2}\\ C=\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{3}{2}\\ C=2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{2}\\ C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{3}{2}\text{ }khi\text{ }x=\dfrac{1}{2}\)

\(D=9x^2-6x+25y^2+10y+4\\ D=9x^2-6x+25y^2+10y+1+1+2\\ D=\left(9x^2-6x+1\right)+\left(25y^2+10y+1\right)+2\\ D=\left[\left(3x\right)^2-2\cdot3x\cdot1+1^2\right]+\left[\left(5y\right)^2+2\cdot5y\cdot1+1^2\right]+2\\ D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \left(5y+1\right)^2\ge0\forall y\\ \Rightarrow\left(3x-1\right)^2+\left(5y+1\right)^2\ge0\forall x;y\\ \Rightarrow D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\ge2\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(3x-1\right)^2=0\\\left(5y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\5y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=1\\5y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{5}\end{matrix}\right.\\ \text{Vậy }D_{\left(Min\right)}=2\text{ khi }x=\dfrac{1}{3};y=-\dfrac{1}{5}\)

Answer 2

[Trần Quốc Lộc]

Câu 2

\(M=x^2+6x+1\\ M=x^2+6x+9-8\\ M=\left(x^2+6x+9\right)-8\\ M=\left(x+3\right)^2-8\\ Do\text{ }\left(x+3\right)^2\ge0\forall x\\ M=\left(x+3\right)^2-8\ge-8\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\\ \text{Vậy }M_{\left(Min\right)}=-8\text{ khi }x=-3\)

\(N=10y-5y^2-3\\ N=10y-5y^2-5+2\\ N=-\left(5y^2-10y+5\right)+2\\ N=-5\left(y^2-2y+1\right)+2\\ N=-5\left(y-1\right)^2+2\\ Do\left(y-1\right)^2\ge0\forall x\\ \Rightarrow-\left(y-1\right)^2\le0\forall x\\ \Rightarrow-5\left(y-1\right)^2\le0\forall x\\ \Rightarrow N=-5\left(y-1\right)^2+2\le2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-1\right)^2=0\\ \Leftrightarrow y-1=0\\ \Leftrightarrow y=1\\ \text{Vậy }N_{\left(Max\right)}=2\text{ khi }y=1\)