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a);b);c) Dùng máy tính (cụ thể là solve) bấm nghiệm rồi phân tích
d)Nhóm số T1;T2;T4 lại vs nhau
e)Biến đổi thành x2-2xy+y2-9y2
a) \(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2+1\right)\)
b) \(4x^2-4x-3=4x^2+2x-6x-3=2x\left(2x+1\right)-3\left(2x+1\right)=\left(2x+1\right)\left(2x+3\right)\)
c) \(4x^4+81=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x-9\right)\)
d) \(x^2-6xy-25+9y^2=\left(x-3y\right)^2-25=\left(x-3y-5\right)\left(x-3y+5\right)\)
e) \(x^2-8y^2-2xy=x^2+2xy-4xy-8y^2=x\left(x+2y\right)-4y\left(x+2y\right)=\left(x+2y\right)\left(x-4y\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
1) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=ab\left(a+b\right)-b^2c-bc^2+a^2c-ac^2\)
\(=ab\left(a+b\right)-c\left(b^2-a^2\right)-c^2\left(a+b\right)\)
\(=ab\left(a+b\right)-c\left(a+b\right)\left(a-b\right)-c^2\left(a+b\right)\)
\(=\left(a+b\right)\left(ab-ac+bc-c^2\right)\)
\(=\left(a+b\right)\left[a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a+b\right)\left(b-c\right)\left(a+c\right)\)
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
mình chỉ làm được câu b) thôi ^^
a^2-2ab+b^2-4x^2
= (a-b)^2-(2x)^2
= (a-b+2x)(a-b-2x)
a/ x2+2x-y2+1=x2+2x+1-y2
= x2+x+x+1-y2
= x(x+1)+(x+1)-y2
= (x+1)2-y2
= (x+1-y)(x+1+y)
b/ a2-2ab+b2-4x2=a2-ab-ab+b2-(2x)2
=a(a-b)-b(a-b)-(2x)2
=(a-b)2-(2x)2
= (a-b-2x)(a-b+2x)
a) ta có: (ab -1)^2 +(a+b)^2 =(ab)^2 -2ab +1 +a^2 +2ab +b^2 = a^2.(b^2 +1)+(b^2 +1)+(2ab-2ab)=(b^2 +1)(a^2 +1)
b)ta có: x^4 +2.x^3 -4x-4=[(x^2)^2 +2.(x^2).x +x^2 ] -(x^2 +2.2.x +4)=(x^2 +x)^2 -(x+2)^2=(x^2 +x+x+2)(x^2 +x-x-2)=(x^2 +2x+2)(x^2 -2)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
a,\(\left(ab-1\right)^2+\left(a+b\right)^2\)
= \(a^2b^2-2ab+1+a^2+2ab+b^2\)
= \(\left(a^2b^2+a^2\right)+\left(b^2+1\right)\)
= \(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
= \(\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^4+2x^3-4x-4\)
= \(\left(x^4-4\right)+\left(2x^3-4x\right)\)
= \(\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
=\(\left(x^2-2\right)\left(x^2+2x+2\right)\)
=