Giải phương trình \(\left(x^2+4x+11\right)\left(y^2-8x^2+21\right)=35\)
Phân tích đa thức thành nhân tử \(a^2\left(b-2c\right)+b^2\left(c-a\right)+2c^2\left(a-b\right)+abc\)
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9 tháng 8 2017
\(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
TM
9 tháng 8 2017
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a\right)+b^2c^2\left(a-c\right)+c^2a^2\left(c-a\right)\)
\(=b^2\left(a-b\right)\left(a^2-c^2\right)+c^2\left(c-a\right)\left(a^2-b^2\right)\)
\(=b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)+c^2\left(c-a\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[-b^2\left(a+c\right)+c^2\left(a+b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(-ab^2-b^2c+ac^2+bc^2\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c^2-b^2\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c-b\right)\left(c+b\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
18 tháng 8 2019
\(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+c^2a^2\left(c-a\right)-b^2c^2\left(c-a\right)\)
\(=\left(a-b\right)b^2\left(a-c\right)\left(a+c\right)+\left(c-a\right)c^2\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(ab^2+cb^2-c^2a-c^2b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)
AH
Akai Haruma
Giáo viên
21 tháng 8 2023
Lời giải:
a.
$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$
$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$
b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?
15 tháng 7 2018
a) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[a^2-\left(b-c\right)^2\right]\)
\(=\left(b+c+a\right)\left(b+c-a\right)\left(a+b-c\right)\left(a-b+c\right)\)
b) \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)
\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)
c) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b+3\right)\left(a-b-3\right)\)
d) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18+4x^2+3x\right)\left(4x^2-3x-18-4x^2-3x\right)\)
\(=\left(8x^2-18\right)\left(-6x-18\right)\)
\(=\left[2\left(4x^2-9\right)\right]\left[-6\left(x+3\right)\right]\)
\(=12\left(2x+3\right)\left(2x-3\right)\left(x+3\right)\)
23 tháng 9 2018
\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right].\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right].\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Tham khảo nhé~
TN
4 tháng 11 2016
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
4 tháng 11 2016
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
