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PA
1 tháng 11 2017
\(M=x^2+xy+y^2-3x-3y\)
\(\Rightarrow4M=4x^2+4xy+4y^2-12x-12y\)
\(=\left(x^2+4y^2+9+4xy-12y-6x\right)+\left(3x^2-6x+3\right)-12\)
\(=\left(x+2y-3\right)^2+3\left(x-1\right)^2-12\ge-12\)
\(\Rightarrow M\ge-3\)
\(\Rightarrow Min_M=-3\Leftrightarrow x=y=1\)
24 tháng 5 2022
Bài 2:
a: \(A=1999\cdot2001\)
\(=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2=B\)
Do đó: B lớn hơn
b: \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}=D\)
Do đó: D lớn hơn
26 tháng 9 2017
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
4 tháng 11 2017
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
TY
4 tháng 11 2017
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
PT
3 tháng 10 2017
\(125-x^6=\left(5\right)^3-\left(x^2\right)^3\)
\(=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)
\(49\left(x-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left[7x-28\right]^2-\left[3y+6\right]^2\)
\(=\left(7x-28-3y-6\right)\left(7x-28+3y+6\right)\)
\(=\left(7x-3y-34\right)\left(7x-22+3y\right)\)
NC
29 tháng 10 2017
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
AN
30 tháng 10 2017
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
30 tháng 5 2022
a: \(a^4+6a^3+11a^2+6a\)
\(=a\left(a^3+6a^2+11a+6\right)\)
\(=a\left(a^3+a^2+5a^2+5a+6a+6\right)\)
\(=a\left(a+1\right)\left(a^2+5a+6\right)\)
\(=a\left(a+1\right)\left(a+2\right)\left(a+3\right)\)
Vì a;a+1;a+2;a+3 là bốn số liên tiếp
nên \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)⋮4!\)
hay \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)⋮24\)
b: \(a^5-5a^3+4a\)
\(=a\left(a^4-5a^2+4\right)\)
\(=a\left(a^2-4\right)\left(a^2-1\right)\)
\(=a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)\)
Vì a;a-2;a+2;a-1;a+1 là 5 số liên tiếp
nên \(a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)⋮5!\)
hay \(a\left(a-2\right)\left(a+2\right)\left(a-1\right)\left(a+1\right)⋮120\)
Ta có:
\(\left(a-1\right)^2\ge0\Leftrightarrow a^2-2a+1\ge0\Leftrightarrow a^2+1\ge2a\) (1)
\(\left(b-1\right)^2\ge0\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow b^2+1\ge2b\) (2)
\(\left(c-1\right)^2\ge0\Leftrightarrow c^2-2c+1\ge0\Leftrightarrow c^2+1\ge2c\) (3)
Từ (1), (2) và (3) suy ra;
\(a^2+1+b^2+1+c^2+1\ge2a+2b+2c\)
<=> \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
<=> \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) \(\xrightarrow[]{}\) đpcm
Dấu "=" xảy ra khi a=b=c=1
Ta có: \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge+2a+2b+2c-3\)
\(\Leftrightarrow a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) (đpcm)
Vậy \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\)