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\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
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)
= (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
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)
a) x4 + 4 = (x4 + 4x2 + 4) - 4x2 = (x2 + 2)2 - 4x2 = (x2 + 2x + 2)(x2 - 2x + 2)
b) (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 x2 + 7x + 10 = y => y(y + 2) - 24 = y2 + 2y - 24
= y2 + 6y - 4y - 24 = (y - 4)(y + 6) = (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16) = (x2 + x + 6x + 6)(x2 + 7x + 16) = (x + 1)(x + 6)(x2 + 7x + 16)
<=>[(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+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)
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)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=y\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)