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9 tháng 8 2019
11.\(8a^3-12a^2+6a-1=\left(3a-1\right)^3\)
12.\(x^3+12x^2+48x+64=\left(x+4\right)^3\)
13.\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
14.\(a^3-b^3+c^3+3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)+c^3+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-bc+ac\right)\)
15.\(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-4x^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
16.\(4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-4x^2y^2\)
\(=\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\)
17.\(x^4+324=x^4+36x^2+18^2-36^2\)
\(=\left(x^2+18\right)^2-36x^2\)
\(=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
18.\(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
19.\(x^{11}+x+1=x^{11}-x^8+x^8-x^5+x^5-x^2+x^2+x+1\)
\(=x^8\left(x^3-1\right)+x^5\left(x^3-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^8\left(x-1\right)\left(x^2+x+1\right)+x^5\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)
20.\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
Đặt \(t=x^2-11x+30\) thay vào phương trình ta được:
\(\left(t-2x\right).t-24x^2\)
\(=t^2-2tx-24x^2\)
\(=t^2+4tx-6tx-24x^2\)
\(=t\left(t+4x\right)-6x\left(t+4x\right)\)
\(=\left(t+4x\right)\left(t-6x\right)\)
\(=\left(x^2-11x+30+4x\right)\left(x^2-11x+30-6x\right)\)
\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
Lời giải:
\(P(x)=x(x+2)(x+3)(x+5)-7\)
\(=[x(x+5)][(x+2)(x+3)]-7\)
\(=(x^2+5x)(x^2+5x+6)-7\)
\(=a(a+6)-7\) (đặt \(x^2+5x=a\) )
\(=a^2+6a-7=a^2-a+7a-7\)
\(=a(a-1)+7(a-1)=(a-1)(a+7)\)
\(=(x^2+5x-1)(x^2+5x+7)\)
-----------------
\(Q(x)=(4x-2)(10x+4)(5x+7)(2x+1)+17\)
\(=4(2x-1)(5x+2)(5x+7)(2x+1)+17\)
\(=4[(2x-1)(5x+7)][(5x+2)(2x+1)]+17\)
\(=4(10x^2+9x-7)(10x^2+9x+2)+17\)
\(=4a(a+9)+17\) (đặt \(10x^2+9x-7=a\)
\(=4a^2+36a+17=(2a+9)^2-8^2\)
\(=(2a+9-8)(2a+9+8)=(2a+1)(2a+17)\)
\(=(20x^2+18x-13)(20x^2+18x+3)\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
\(R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2\)
\(=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2\)
\(=(9x^2-9x-10)(9x^2+x-10)+24x^2\)
\(=(a-9x)(a+x)+24x^2\) (đặt \(9x^2-10=a\) )
\(=a^2-8ax+15x^2=(a^2-5ax)-(3ax-15x^2)\)
\(=a(a-5x)-3x(a-5x)=(a-3x)(a-5x)\)
\(=(9x^2-3x-10)(9x^2-5x-10)\)
--------------------------
\(H(x)=(x-18)(x-7)(x+35)(x+90)-67x^2\)
\(=[(x-18)(x+35)][(x-7)(x+90)]-67x^2\)
\(=(x^2+17x-630)(x^2+83x-630)-67x^2\)
\(=a(a+66x)-67x^2\) (đặt \(x^2+17x-630=a\) )
\(=a^2-ax+67ax-67x^2\)
\(=a(a-x)+67x(a-x)=(a-x)(a+67x)\)
\(=(x^2+16x-630)(x^2+84x-630)\)
17) \(\left(x^2-11x+30\right)\left(x^2-13x+30\right)=24x^2\)
\(\left(x-11+\frac{30}{x}\right)\left(x-13+\frac{30}{x}\right)=24\)
\(t\left(t-2\right)=24\)
\(\left(t-1\right)^2=25\)
t =6 hoặc t =-4
+\(\left(x-11+\frac{30}{x}\right)=6\Leftrightarrow x^2-11x+30=6x\Leftrightarrow x^2-17x+30=0\)
+\(\left(x-11+\frac{30}{x}\right)=-4\)