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Làm dễ hiểu chút
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100-99\right)\left(99+100\right)\)
\(=3+7+...+199\)
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(=21^8-\left(21^8-1\right)=1\)
Vậy A > B
a)x3-7x+6
=x3+0x2-7x+6
=x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2-2x+3x-6)
=(x-1)[x(x-2)+3(x-2)]
=(x-1)(x+3)(x-2)
bạn hỏi từng câu 1 lần thôi cũng đc hỏi 1 lần 17 câu thì thánh nào vô kiên nhẫn trả lời hết đc ^^
a, 3x^2 + 13x + 10
= 3x^2 + 3x + 10x + 10
= 3x(x + 1) + 10(x + 1)
= (3x + 10)(x + 1)
b, x^2 - 10x + 21
= x^2 - 3x - 7x + 21
= x(x - 3) - 7(x - 3)
= (x - 7)(x - 3)
c, 6x^2 - 5x + 1
= 6x^2 - 3x - 2x + 1
= 3x(2x - 1) - (2x - 1)
= (3x - 1)(2x - 1)
Bạn đăng 1 lần nhiều bài như vậy làm người khác nản lắm đấy =) đơn giản bài rất dài mà mik cx ko chắc là bản thân mik có đc k hay ko nên phải nản vậy thôi :)
1a)\(3x^2+13x+10=3x^2+3x+10x+10\)
\(3x\left(x+1\right)+10\left(x+1\right)=\left(3x+10\right)\left(x+1\right)\)
b)\(x^2-10x+21=x^2-3x-7x+21\)
\(=x\left(x-3\right)-7\left(x-3\right)=\left(x-7\right)\left(x-3\right)\)
c)\(6x^2-5x+1=6x^2-3x-2x+1\)
\(=3x\left(2x-1\right)-\left(2x-1\right)=\left(3x-1\right)\left(2x-1\right)\)
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(A=2^2-1^2+4^2-3^2+...+100^2-99^2\)
\(A=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)
\(A=1\left(1+2\right)+1\left(3+4\right)+....+1\left(99+100\right)\)
\(A=1+2+3+4+....+99+100\)
A=5050
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(B=\left(3.7\right)^8-\left(21^8-1\right)\)
\(B=21^8-21^8+1\)
B=1
mà A=5050
⇒ A>B