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a) \(\left(-12x^{13}y^{15}+6x^{10}y^{14}\right):\left(-3x^{10}y^{14}\right)\)
\(=-12x^{13}y^{15}:-3x^{10}y^{14}+6x^{10}y^{14}:-3x^{10}y^{14}\)
\(=4x^3y-2\)
b) \(\left(x-y\right)\left(x^2-2x+y\right)-x^3+x^2y\)
\(=x^3-2x^2+xy-x^2y+2xy-y^2-x^3+x^2y\)
\(=-2x^2+3xy-y^2\)
a) \(-12x^{13}\)\(y^{15}\)+\(6x^{10}\)\(y^{14}\):\(-3x^{10}\)\(y^{14}\)
=\(-12x\)\(^{13}\)\(y^{15}\)\(:\)\(-3x^{10}y^{14}\)\(+6x^{10}y^{14}:-3x^{10}y^{14}\)
\(=4x^3y-2\)
b)\(=\left(x-y\right)x^2-2x+y-x^3+x^2y\)
\(=x^3-x^2y-2x+y-x^3+x^2y\)
\(=-2x+y\)
Bài 1:
\(\left(2x+3\right)^2+\left(2x-3\right)^2+2\left(2x+3\right)\left(2x-3\right)\)
\(=\left(2x+3+2x-3\right)^2=\left(4x\right)^2=16x^2\)
Bài 2:
a, \(\left(x^2+xy+y^2\right)\left(x-y\right)+\left(x^2-xy+y^2\right)\left(x+y\right)\)
\(=x^3-y^3+x^3+y^3=2x^3\)
b, \(\left(2a-b\right)\left(4a^2+2ab+b^2\right)\)
\(=\left(2a\right)^3-b^3=8a^3-b^3\)
c, \(13x\left(3-x\right)-12\left(x+1\right)\)
\(=39x-13x^2-12x-12=-13x^2-27x-12\)
d, \(\left(2x-1\right)\left(x+12\right)\left(x^2+14\right)\)
\(=\left(2x^2+24x-x-12\right)\left(x^2+14\right)\)
\(=2x^4+23x^3-12x^2+28x^2+322x-168\)
\(=2x^4+23x^3+16x^2+322x-168\)
e, Giống câu b
Chúc bạn học tốt!!!
a,\(=x^3+x^2-\left(31x^2+31x\right)\)
\(=x^2\left(x+1\right)-31x\left(x+1\right)\)
\(=\left(x^2-31x\right)\left(x+1\right)=\left(31^2-31^2\right)\left(31+1\right)=0\)
b, Phân tích 3 số hạng đầu ta có:\(=x^5-x^4-\left(14x^4-14x^3\right)=\left(x^4-14x^3\right)\left(x-1\right)=\left(14^4-14^4\right)\left(x-1\right)=0\)
Thay x= 14 vào ta có: \(-29.14^2+13.14=-5502\)
c, do x=9 => x+1=10; Thay vào ta có:
\(C=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(C=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-....+x^3+x^2-x^2-x+10\)
\(C=-x+10=-9+10=1\)
CHÚC BẠN HỌC TỐT.....
9) \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=b^2\left[a^2+2ab+b^2+a\left(a-b\right)+b\left(a-b\right)+a^2-2ab+b^2\right]\)
\(=b^2\left(a^2+2ab+b^2+a^2-ab+ab-b^2+a^2-2ab+b^2\right)\)
\(=b^2\left(3a^2+b^2\right)\)
10) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1-3x-2\right)\left(6x-1+3x+2\right)\)
\(=\left(3x-3\right)\left(9x+1\right)\)
11) \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
12) \(\left(x^2-25\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-25-x+5\right)\left(x^2-25+x-5\right)\)
\(=\left(x^2-x-20\right)\left(x^2-30+x\right)\)
13) \(x^6-x^4+2x^3+2x^2\)
\(=x^6-x^4+2x^3+2x^2-1+1\)
\(=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left[\left(x^3\right)^2+2x^3.1+1^2\right]-\left[\left(x^2\right)^2-2x^2.1+1^2\right]\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2\)
\(=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)\)
\(=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
1) \(\left(x+y\right)^2-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
2) \(100-\left(3x-y\right)^2\)
\(=10^2-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
3) \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
4) \(4a^2b^4-c^4d^2\)
\(=\left(2ab^2\right)^2-\left(c^2d\right)^2\)
\(=\left(2ab^2-c^2d\right)\left(2ab^2+c^2d\right)\)
5) Đề đúng ko vậy ạ?
6) \(16x^3+54y^3\)
\(=2\left(8x^3+27y^3\right)\)
\(=2\left[\left(2x\right)^3+\left(3y\right)^3\right]\)
\(=2\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]\)
\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
7) \(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2xy+y^2\right]\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
8) \(\left(a+b\right)^2-\left(2ab-b\right)^2\)
\(=\left(a+b-2ab+b\right)\left(a+b+2ab-b\right)\)
\(=\left(a+2b-2ab\right)\left(a+2ab\right)\)
Đáp án cần chọn là: D