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, bạn nào biết thì giúp mình sửa nhé
a, Theo bài ra ta có:
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
b, theo bài ra ta có:
\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)
\(=x^2\left(x-3\right)-2x\left(x-3\right)-3\left(x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x-3\right)\)
c,Theo bài ra ta có:
\(=x^3+5x^2+3x^2+15x+2x+10\)
\(=x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2+3x+2\right)\)
\(=\left(x+5\right)\left(x^2+x+2x+2\right)=\left(x+5\right)\left(x\left(x+1\right)+2\left(x+1\right)\right)\)
\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)
CHÚC BẠN HỌC TỐT...........
a) \(x^3-3x+2\)
= \(x^3-x^2+x^2-x-2x+2\)
= \(x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x-2\right)\)
= \(\left(x-1\right)\left(x^2+2x-x-2\right)\)
= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(x-1\right)\)
= \(\left(x-1\right)^2\left(x+2\right)\)
b) \(x^3-5x^2+3x+9\)
= \(x^3+x^2-6x^2-6x+9x+9\)
= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-6x+9\right)\)
= \(\left(x+1\right)\left(x-3\right)^2\)
c) \(x^3+8x^2+17x+10\)
= \(x^3+x^2+7x^2+7x+10x+10\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10\right)\)
= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)
= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
d) \(x^3-3x^2+6x+4\)
Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:
\(x^3+3x^2+6x+4\)
= \(x^3+x^2+2x^2+2x+4x+4\)
= \(x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+2x+4\right)\)
Học tốt nhé :))
Bài 1:
a, \(2x\left(y-z\right)+5y\left(z-y\right)=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b, \(x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1\)
\(=x^2.\left(x-1\right)-2x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)=\left(x-1\right)\left(x^2-x-x+1\right)\)
\(=\left(x-1\right)\left(x-1\right)^2=\left(x-1\right)^3\)
c, \(7x^2-7xy-4x+4y=7x.\left(x-y\right)-4.\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d, \(x^2-6x+8=x^2-2x-4x+8\)
\(=x.\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)
Chúc bạn học tốt!!!
1)
a) \(2x\left(y-z\right)+5y\left(z-y\right)\)
\(=2x\left(y-z\right)-5y\left(y-z\right)\)
\(=\left(y-z\right)\left(2x-5y\right)\)
b) \(x^3-3x^2+3x-1\)
\(=x^3-3.x^2.1+3.x.1^2-1^3\)
\(=\left(x-1\right)^3\)
c) \(7x^2-7xy-4x+4y\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
d) \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
2)
a) \(\left(5x^2+3x-1\right)\left(x+3\right)\)
\(=5x^3+3x^2-x+15x^2+9x-3\)
\(=5x^3+3x^2+15x^2-x+9x-3\)
\(=5x^3+18x^2+8x-3\)
b) \(\left(x^3+2x^2+3x-1\right):\left(x^2-2\right)\)
\(=x+2+\dfrac{5x+3}{x^2-2}\)
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)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Xin mọi người ai biết giúp mình nha. Mình cảm ơn nhiều. Tí nữa mình đi học rồi
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)....\left(2^{64}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)....\left(2^{64}+1\right)+1\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)......\left(2^{64}+1\right)+1\)
\(A=\left(2^8-1\right)\left(2^8+1\right)......\left(2^{64}+1\right)+1\)
\(A=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(A=2^{128}-1+1=2^{128}\)
Ribi Nkok Ngok hãy cảm nhận sự khác biệt cách tiếp cận sử lý bài toán (B là huyền thoại)\(B=1^2-2^2+3^2-4^2+....-2010^2+2011^2\)
\(\text{B=1+(3-2)(3+2) +(5-4)(5+4)+....+(2011-2010)(2011+2010)}\)\(\text{B=1+(3+2) +(5+4)+....+(2011+2010)}\)\(B=1+2+3+...+2011\)
\(B=\dfrac{2011.2012}{2}=2011.1006=2023066\)
Lời giải
\(\left(x^2+x\right)^2+\left(x^2+x\right)=y^2-3\)
\(\left(2x^2+2x+1\right)^2=4y^2-11\)
\(\Leftrightarrow Z^2-P^2=11\Rightarrow\left\{{}\begin{matrix}Z^2=36\\P^2=25\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=\pm3\\2x^2+2x+1=\pm5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\pm3\\\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\end{matrix}\right.\)
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
= (x + y)3 + z3 – 3x2y – 3xy2 - 3xyz
= (x + y +z)[(x + y)2 – (x + y)z + z2)] - 3xy(x + y + z)
= (x + y + z)(x2 +2xy + y2 – xz – yz +z2 – 3xy)
= (x + y + z)(x2 + y2 +z2 – xy - yz – xz)
2(x4+y4+z4)-(x2+y2+z2)2-2(x2+y2+z2)(x+y+z)2+(x+y+z)4
=2(x4+y4+z4)-(x2+y2+z2)2+(x+y+z)2[-2(x2+y2+z2)+(x+y+z)2]
tới đây r` sao đặt ẩn phân tích tiếp chắc =="
a) 8 - 27x3
= (2 - 3x) (4 + 6x + 9x2)
b) 5x2y - 10xy - 5y2
= 5y(x2 - 2x - y)
c) x3 + 8y3
= (x + 2y) (x2 - 2xy + 4y2)
d) 3x2 + 5y - 3xy - 5x
= (x - y) (3x - 5)