Phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4) – 24
phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4)-24
= (x+1)(x+4)(x+2)(x+3)-24
= (x2 +5x+4) (x2 +5x+6)-24
Đặt x2 +5x+4 =a
=>(x2 +5x+4)(x2+5x+6)-24
= a(a+2)-24 = a2 +2a-24
= a2 +6a-4a-24
= a(a+6) - 4(a+6) = (a-4)(a+6)
= (x2 +5x+a-4)(x2 +5x+4+6) = (x2 +5x)(x2 +5x+10)
Phân tích đa thức thành nhân tử:
(x + 1)(x + 2)(x + 3)(x + 4) - 24
=(x+1)(x+4)(x+2)(x+3) - 24
=(x^2+5x+4)(x^2+5x+6) - 24
=(x^2+5x+5-1)(x^2+5x+5+1) - 24 [hằng đẳng thức a^2-b^2 nha]
=(x^2+5x+5)^2-1^2-24
=(x^2+5x+5)^2 - 25
=(x^2+5x+5)^2 - 5^2
=(x^2+5x+5-5)(x^2+5x+5+5)
=(x^2+5x)(x^2+5x+10
Phân tích đa thức thành nhân tử
( x + 1 ) (x + 2) (x+3) (x+4) - 24
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=t\left(t+2\right)-24\)
\(=t^2+2t+1-25\)
\(=\left(t+1\right)^2-5^2\)
\(=\left(t+1+5\right)\left(t+1-5\right)\)
\(=\left(t+6\right)\left(t-4\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
Phân tích đa thức thành nhân tử:
( x + 1)( x+ 2)( x + 3)( x + 4) - 24
Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\)\(\left(x+4\right)-24\)
= \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) (*)
. Đặt \(x^2+5x+4=t\) (1)
(*) <=> \(t\left(t+2\right)-24=t^2+2t-24=\left(t-4\right)\left(t+6\right)\) (2)
Thay (1) vào (2) ta suy ra : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\) \(\left(x+4\right)-24=\)\(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\) = \(\left(x^2+5x\right)\left(x^2+5x+10\right)\) = \(x\left(x+5\right)\left(x^2+5x+10\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(=\left(x^2+5x+4\right)^2+2.\left(x^2+5x+4\right)+1-25\)
\(=\left(x^2+5x+4+1\right)^2-5^2\)
\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
Ta có
<=>(x+1)(x+4)(x+2)(x+3)-24
<=>(X^2+5x4)(x^2+5x+6)-24
Đặt x^2+5x+5=x (1)
Ta có
<=>(x+1)(x-1)-24
<=>x^2-25
Thay 1 vào x ta có
(x^2+5x+5)^2-5^2
<=>(x^2+10)(x^2+5x)(dpcm)
phân tích đa thức thành nhân tử :
(x + 2) (x + 3) (x + 4) (x + 5) - 24
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25\)
\(=\left(x^2+7x+11\right)^2-25=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
=[(x+2)(x+5)][(x+3)(x+4)]−24=(x2+7x+10)(x2+7x+12)−24=(x2+7x+11)2−1−24=(x2+7x+11)2−25=(x2+7x+11−5)(x2+7x+11+5)=(x2+7x+6)(x2+7x+16)=(x+1)(x+6)(x2+7x+16)
phân tích đa thức thành nhân tử
(x+1)(x+2)(x+3)(x+4)-24
(x+1)(x+4)(x+2)(x+3)-24
=(x2+5x+4)(x2+5x+6)-24
=(x2+5x+5-1)(x2+5x+5+1)-24
=(x2+5x+5)2-1-24
=(x2+5x+5)2-25
=x(x2+5x+10)(x+5)
Nhân tử là gì bạn ơi
giờ này còn đi hỏi bài làm gì
Sao em không tự làm đi
Đã ngu đã giốt còn hay hỏi nhiều
( x + x + x + x ) + ( 1+ 2+3+4 ) -24
4x + 10 -24
đến đây thì chịu
phân tích đa thức thành nhân tử
(x+1)(x+2)(x+3)(x+4) - 24
Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(\left(x^2+5x+4\right)^2+2.\left(x^2+5x+4\right)+1-25\)
\(=\left(x^2+5x+5\right)-5^2\)
\(=x\left(x+5\right)\left(x^2+5x-10\right)\)
M=(x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+5 là a (1)
Từ 2 đk trên=>M=(a-1)(a+1)-24
=>M=a^2 - 1-24
=a^2-25
=(a-5)(a+5) và (1)
=(x^2+5x+5-5)(x^2+5x+5+5)
=(x^2+5x)(x^2+5x+10)
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(x+ 2)(x +3)(x+ 4)(x+ 5) -24 -> phân tích đa thức sau thành nhân tử?
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=y\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+16\right)\left(x^2+7x+6\right)\\ =\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
Phân tích đa thức thành nhân tử: (x+2)(x+3)(x+4)(x+5) - 24
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\) ta có:
\(=t\left(t+2\right)-24=t^2+2t-24\)
\(=t^2-4t+6t-24\)\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
(x+2)(x+3)(x+4)(x+5)-24
=(x^2+7x+10)(x^2+7x+12)-24
Đặt x^2+7x+10=a
a(a+2)-24
=a^2+2a-24
=(a-4)(a+6)
=(x^2+7x+6)(x^2+7x+16)
=(x+1)(x+6)(x^2+7x+16)