phân tích đa thức thành nhân tử: (x+2)(x+3)(x+4)(x+5)-24
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)
(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)
Phân tích đa thức thành nhân tử (x+2)(x+3)(x+4)(x+5)-24
= (x+2)(x+5)(x+3)(x+4)-24
= (x2+7x+10)(x2+7x+12)-24
đặt y=x2+7x+10
ta có biểu thức:
y.(y+2)-24
= y2+2y-24
= y2+6y-4y-24
= y(y+6)-4(y+6)
= (y+6)(y-4)
= (x2+7x+10+6)(x2+7x+10-4)
= (x2+7x+16)(x2+7x+6)
(x + 2)(x + 3)(x + 4)(x + 5) - 24
= (x2 + 3x + 2x + 6)(x2 + 5x + 4x + 20) - 24
= (x2 + 5x + 6)(x2 + 9x + 20) - 24
= x4 + 9x3 + 20x2 + 5x3 + 45x2 + 100x + 6x2 + 54x + 120 - 24
= x4 + 14x3 + 71x2 + 100x + 96
(x + 2)(x + 3)(x + 4)(x + 5) – 24 = (x + 2)(x + 5)(x + 4)(x + 3) – 24
= ( x2 + 7x +10) ( x2 + 7x +12) – 24 đặt x2 + 7x +10 = a
= a(a + 2 ) – 24 = a2 + 2a – 24 = a2 + 2a +1 – 25
= (a – 4 )( a + 6)
Thay vào ta có: (x + 2)(x + 3)(x + 4)(x + 5) – 24 = ( x2 + 7x +6) ( x2 + 7x +16)
= (x + 6)(x + 1)( x2 + 7x +16)
phân tích đa thức thành nhân tử (x+2)(x+3)(x+4)(x+5)-24
Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và x^2+5x+5=a
Do đó A= (a-1)(a+1)-24
= a^2- 25
= a^2-5^2
=(a-5)(a+5)
= ( x^2+5x+5-5)( x^2+5x+5+5)
= ( x^2+5x)(x^2+5x+10)
Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và x^2+5x+5=a
Do đó A= (a-1)(a+1)-24
= a^2- 25
= a^2-5^2
=(a-5)(a+5)
= ( 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+2)(x+3)(x+4)(x+5)-24
<=>[(x+2)(x+5)][(x+3)(x+4)]-24=\(\left(x^2+7x+10\right)\left(\left(x^2+7x+12\right)\right)-24\)(1)
đặt x^2+7x+11=t
=> (1)<=> (t-1)(t+1)-24=t^2-1-24=t^2-25=(t-5)(t+5)
<=> \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
(x+2)(x+3)(x+4)(x+5)=24
(x+x+x+x)(2+3+4+5)=24
(x.4)14=24
x.4=24:14
x.4=2
x=2:4
X=1/2
(X+2)(X+3)(X+4)(X+5)-24
phân tích đa thức thành nhân tử
(x+2)(x+5)(x+4)(x+3)-24=(x^2+7x+10)(x^2+7x+12)-24
đặt:x^2+7x+10=t thi x^2+7x+12=t+2
=>t(t+2)-24=t^2+2t-25=t^2+2t+1-25=(t+1)^2-5^2=(t-4)(t+6)
thay t vao suy ra: (x^2+7x+6)(x^2+7x+16)
\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
thang: Em sai một chỗ rất nhỏ :) tách \(t^2+2t-24\) mới đúng nhé :)
Phân tích đa thức thành nhân tử:
(x+2)(x+3)(x+4)(x+5)-24
ta có: A= (x+2)*(x+3)*(x+4)*(x+5)=\(\left(\left(x+2\right)\cdot\left(x+5\right)\right)\cdot\left(\left(x+3\right)\cdot\left(x+4\right)\right)=\left(x^2+7x+10\right)\cdot\left(x^2+7x+12\right)\)
đặt t=x^2+7x+11
=> A=(t-1)*(t+1)-24=t^2-25=(t-5)*(t+5)
=> A=(x^2+7x+6)*(x^2+7x+16)
Phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4) – 24
Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
- Đặt \(x^2+5x+5=a\)
\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)