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a:Sửa đề: \(\left(2x+5\right)^2-\left(x-3\right)^2\)
=(2x+5-x+3)(2x+5+x-3)
=(x+8)(3x+2)
b:Sửa đề: \(25\left(2x-1\right)^2-9\left(x+1\right)^2\)
\(=\left(10x-5\right)^2-\left(3x+3\right)^2\)
=(10x-5-3x-3)(10x-5+3x+3)
=(7x-8)(13x-2)
c: \(1-9x+27x^2-27x^3\)
\(=1^3-3\cdot1^2\cdot3x+3\cdot1\cdot\left(3x\right)^2-\left(3x\right)^3\)
\(=\left(1-3x\right)^3\)
d: \(49-a^2+2ab-b^2\)
\(=7^2-\left(a-b\right)^2\)
=(7-a+b)(7+a-b)
e: \(-4x^2-12xy-9y^2+25\)
\(=25-\left(4x^2+12xy+9y^2\right)\)
\(=5^2-\left(2x+3y\right)^2\)
=(5-2x-3y)(5+2x+3y)
Từ điểm B, C vẽ các đường thẳng lần lượt đi qua AC và AB và cắt AC tại D, AB tại E. Sao cho BE = DC.
Xét tam giác BEC và tam giác DCB có:
BE = DC ( chứng minh trên )
ˆB=ˆC( giả thiết )
Cạnh BC chung
=> Tam giác BEC = tam giác DCB ( c.g.c )
Vậy nếu ˆB=ˆCthì AB = AC ( đpcm )
x³ -7x +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-2)(x+3)
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
a) bt \(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x+1\right)\left(x-2\right)\)
kl: ...
b) \(=\left(x+2\right)\left(x^2-8x-15\right)=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
kl:....
a, \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x^2-2x+x-2\right)\)
\(=\left(x-8\right)\left[x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b, \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left(x^2-3x+2x-6\right)\)
\(=\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
Chúc bạn học tốt!!!
b) \(9x^3+6x^2+x\)
\(=x\left(9x^2+6x+1\right)\)
\(=x\left(3x+1\right)^2\)
c) \(x^4+5x^3+15x-9\)
\(=\left(x^4-9\right)+5x\left(x^2+3\right)\)
\(=\left(x^2-3\right)\left(x^2+3\right)+5x\left(x^2+3\right)\)
\(=\left(x^2+3\right)\left(x^2-3+5x\right)\)
a) \(x^2-y^2+10y-25\)
\(=x^2-\left(y^2-10y+25\right)\)
\(=x^2-\left(y-5\right)^2\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
=x3(x+2)-13x2+12x-26x+24
=x3(x+2)-x(13x-12)-2(13x-12)
=x3(x+2)-(13x-12)(x+2)
=(x+2)(x3-x-12x+12)
(x+2)[(x2-1)-12(x-1)]
=(x+2)[x(x-1)(x+1)-12(x-1)]
=(x+2)(x-1)[x(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x(x-3)+4(x+3)]
=(x+2)(x-1)(x-3)(x+4)
trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!
\(a,\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)
\(b,27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
\(c,x^6-x^6\)
\(=0\)
\(d,10x-25-x^2\)
\(=-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
Giải:
Đặt y = x^2 + x
Khi đó, đa thức trở thành:
xy^2 - 2y - 15
=xy^2 - 5y + 3y -15
= y(xy - 5) + 3(xy -5)
= (y+ 3)(xy -5)
Thay y vào, ta được:
(x^2 - x + 3)[x(x^2 - x) - 5]
=(x^2 - x + 3)(x^3 - x^2 - 5)
Sửa đề: \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
a. 3x2– 7x + 2 = 3x2 – 6x – x + 2
= 3x(x -2) – (x - 2)
= (x - 2)(3x - 1)
b. a(x2 + 1) – x(a2 + 1) = ax2 + a – a2x – x
= ax(x - a) – (x - a)
= (x - a)(ax - 1)
a) \(3x^2-7x+2=3x^2-x-6x+2=x\left(3x-1\right)-2\left(3x-1\right)=\left(3x-1\right)\left(x-2\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)=\left(a^2+1\right)\left(a-x\right)\)
a) \(x^2+6x+9\)
\(=\left(x+3\right)^2\)
\(=\left(x+3\right)\left(x+3\right)\)
b) \(10x-25-x^2\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
\(=-\left(x-5\right)\left(x-5\right)\)
c) \(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(\frac{1}{25}x^2-64y^2\)
\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)
b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)
c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)
d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)
\(=\left(x+3-5\right)\left(x+3+5\right)=\left(x-2\right)\left(x+8\right)\)
\(=\left(x+3-5\right)\left(x+3+5\right)=\left(x-2\right)\left(x+8\right)\)