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Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
\(1,\\ A=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)
Đề sai, biểu thức A ko có m thì sao chứng minh?
\(2,\) Gọi 2 số nguyên lt là \(a;a+1\left(a\in Z\right)\)
Ta có \(a+1-a=1\) là số lẻ (đpcm)
\(3,P=9x^2+24x+16-10x-x^2+16=8x^2+14x+32\)
\(4,Q=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x-2=0\Leftrightarrow x=2\)
a.
\(A=x^2-4x+4+2=\left(x-2\right)^2+2\ge2\)
GTNN của A đạt 2 khi và chỉ khi \(x=2\)
b.
\(B=y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
GTNN của B đạt \(\dfrac{3}{4}\) khi và chỉ khi \(y=\dfrac{1}{2}\)
c.
\(C=x^2-4x+4+y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
GTNN của C đạt \(\dfrac{3}{4}\) khi và chỉ khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
a) \(A=x^2-4x+6\)
\(A=x^2-4x+4+2\)
\(A=\left(x-2\right)^2+2\)
Mà: \(\left(x-2\right)^2\ge0\forall x\) nên \(A=\left(x-2\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra:
\(\left(x-2\right)^2+2=2\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(A_{min}=2\) khi \(x=2\)
b) \(B=y^2-y+1\)
\(B=y^2-2\cdot\dfrac{1}{2}\cdot y+\dfrac{1}{4}+\dfrac{3}{4}\)
\(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left(y-\dfrac{1}{2}\right)^2\ge\forall x\) nên \(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow y-\dfrac{1}{2}=0\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\)
c) \(C=x^2-4x+y^2-y+5\)
\(C=x^2-4x+4+y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\)
\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\end{matrix}\right.\) nên
\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}x-2=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(C_{min}=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
`B = x^2- 2xy + y^2 + 2x - 10y + 17
`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`
`= (x-y)^2 + (x+2)^2 + (y-5)^2 + 5 >= 5`.
a: \(M=2x^2-4x+3\)
\(=2x^2-4x+2+1\)
\(=2\left(x^2-2x+1\right)+1\)
\(=2\left(x-1\right)^2+1>=1\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
b: \(N=x^2-4x+5+y^2+2y^2\)
\(=x^2-4x+4+3y^2+1\)
\(=\left(x-2\right)^2+3y^2+1>=1\forall x,y\)
Dấu '=' xảy ra khi x-2=0 và y=0
=>x=2 và y=0
\(A=2n^2\left(2n-1\right)-3\left(2n-1\right)+2=\left(2n^2-3\right)\left(2n-1\right)+2\)
Do \(\left(2n^2-3\right)\left(2n-1\right)⋮2n-1\)
\(\Rightarrow2⋮2n-1\)
\(\Rightarrow2n-1=Ư\left(2\right)\)
Mà 2n-1 luôn lẻ \(\Rightarrow2n-1=\left\{-1;1\right\}\)
\(\Rightarrow n=\left\{0;1\right\}\)
2.
\(Q=-\left(x^2+4x+4\right)-\left(y^2-2y+1\right)+7\)
\(Q=-\left(x+2\right)^2-\left(y-1\right)^2+7\le7\)
\(Q_{max}=7\) khi \(\left(x;y\right)=\left(-2;1\right)\)
\(A=x^2+y^2+\left(\dfrac{1}{2}\right)^2-2xy+2.\dfrac{1}{2}x-2.\dfrac{1}{2}.y+\dfrac{3}{4}\)
\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)
\(Q=x^2+y^2-4x-y+7\)
\(=x^2-4x+4+y^2-y+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x,y\)
Dấu '=' xảy ra khi x=2 và \(y=\dfrac{1}{2}\)
cảm ơn chị ạ