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\(\begin{array}{l}\left( {5{x^2}} \right).\left( {3{x^2} - x - 4} \right)\\ = 5{x^2}.3{x^2} - 5{x^2}.x - 5{x^2}.4\\ = 15{x^4} - 5{x^3} - 20{x^2}\end{array}\)
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
\(\begin{array}{l}\left( {2x + 3} \right).\left( {{x^2} - 5x + 4} \right)\\ = 2x.\left( {{x^2} - 5x + 4} \right) + 3.\left( {{x^2} - 5x + 4} \right)\\ = 2x.{x^2} - 2x.5x + 2x.4 + 3{x^2} - 3.5x + 3.4\\ = 2{x^3} - 10{x^2} + 8x + 3{x^2} - 15x + 12\\ = 2{x^3} + \left( { - 10{x^2} + 3{x^2}} \right) + \left( {8x - 15x} \right) + 12\\ = 2{x^3} - 7{x^2} - 7x + 12\end{array}\)
a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
Rút gọn thôi chứ phân tích sao được ._.
( x - 3 )2 - ( 4x + 5 )2 - 9( x + 1 )2 - 6( x - 3 )( x + 1 )
= x2 - 6x + 9 - ( 16x2 + 40x + 25 ) - 9( x2 + 2x + 1 ) - 6( x2 - 2x - 3 )
= x2 - 6x + 9 - 16x2 - 40x - 25 - 9x2 - 18x - 9 - 6x2 + 12x + 18
= -30x2 - 52x - 7
Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))
Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)
\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)
\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)
\(=\left(4x+7\right)\left(12x+17\right)\)
đặt \(x^2+4x+8=a\)
=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(a+x\right)\left(a+2x\right)\)
b) ta có
\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
đặt \(x^2+8x+11=a\)
=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)
\(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)
Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)
\(A=y^2+2y+1-25\)
\(A=\left(y+1\right)^2-5^2\)
\(A=\left(y+1-5\right)\left(y+1+5\right)\)
\(A=\left(y-4\right)\left(y+6\right)\)
\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
a) Ta có:
\(\begin{array}{l}\left( {x + 1} \right).\left( {{x^2} - x + 1} \right)\\ = {x^3} - {x^2} + x + {x^2} - x + 1\\ = {x^3} + \left( {{x^2} - {x^2}} \right) + \left( {x - x} \right) + 1 = {x^3} + 1\end{array}\)
b) Quy tắc nhân hai đa thức trong trường hợp một biến: ta lấy đơn thức của đa thức này nhân với từng đơn thức của đa thức kia rồi cộng các kết quả với nhau.