To factorize the expression x2+1x2−4(x+1x)+6, we can follow these steps: Step 1: Rewrite the expressionStart by rewriting the expression for clarity:x2+1x2−4(x+1x)+6 Step 2: Use the identity for a2+b2Recall the identity:a2+b2=(a+b)2−2abLet a=x and b=1x. Then...
Factorize: a x+b x+a y+b y, a x^2+b y^2+b x^2+a y^2, a^2+b c+a b+a c, a x-a y+b x-b y
@@ -1,10 +1,14 @@ name: Python 3.9 (x64) name: Python Check on: [push] jobs: build: check: name: Check Python ${{ matrix.python_version }} strategy: fail-fast: false matrix: python_version: ["3.9", "3.10", "3.11"] runs-on: windows-latest defaults: @@ -16,10 +20,10 ...
Method 1: Original expression=(x^8+x^7+x^6)+(x^5+x^4+x^3)+(x^2+x+1)-(x^7-x^6-x^5)-(x^4-x^3-x^2)=(x^2+x+1)(x^6-x^5+x^3-x^2+1)Method 2: Original expression=(x^8-x^5)+(x^5-x^2)+(x^2+x+1) =(x^2+x+1)(x^6-x^5+x^3-x^2+1)反馈...
题目 【题目】Factorize:(1) x^4-x^3-2x^2-3x-1(2) x^4+x^3+2x^2+5x+3 答案 【解析】(1) (x^2-2x-1)(x^2+x+1)(2) (x+1)^2(x^2-x+3)相关推荐 1Factorize:(1) x4−x3−2x2−3x−1(2) x4+x3+2x2+5x+3 2【题目】Factorize:(1) x^4-x^3-2x^2-3x-1(2) ...
⑴分解因式:x4+2x3+3x2+2x+1;⑵分解因式:x4+7x3+14x2+7x+1;⑶分解因式:x4+y4+(x+y)4. 答案 ⑴法一:原式=x4+2x3+3x2+2x+1=x4+2x3+x2+2x2+2x+1=(x2+x)2+2(x2+x)+1=(x2+x+1)2;法二:原式=(x4+1)+2x(x2+1)+3x2=(x2+1)2−2x2+2x(x2+1)+3x2=...
(2) x3+bx2+ax+ab. (3) 4a2−b2+c2−9d2+4ac+6bd. (4) (a+b)2+(a+c)2−(c+d)2−(b+d)2. 相关知识点: 试题来源: 解析 (1) (1−b)2(a−1). (2) (x+b)(x2+a). (3) (2a+c+b−3d)(2a+c−b+3d). (4) 2(a+b+c+d)(a−d).反馈...
So:2.(b)2r(r-s)²-(s-r)³=(s-r)²[2r-(s-r)]=(s-r)²(3r-s)Solve:3.(a)4x²-4x+1=(2x-1)²So:3.(b)(x+3y)²-6(x+3y)(x-y)+9(x-y)²=[(x+3y)-3(x-y)]²=4(3y-x)²Solve:4.(a)-m²...
(4) x4+x3+94x2+x+1相关知识点: 试题来源: 解析 Shown in the analysis. (1) Original expression=x4+x2y2+14y4−x2y2=(x2+12y2)2−(xy)2=14(2x2+2xy+y2)(2x2−2xy+y2) (2) Original expression=x4+8x2y2+16y4−9x2y2=(x2+4y2)2−(3xy)2=(x2+3xy+4y2)...
DWTELEM *b3 = buffer + mirror(-4 + 2, height - 1) * stride; DWTELEM *b0 = buffer + avpriv_mirror(-4 - 1, height - 1) * stride; DWTELEM *b1 = buffer + avpriv_mirror(-4, height - 1) * stride; DWTELEM *b2 = buffer + avpriv_mirror(-4 + 1, height - 1) * stride...