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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(x^2\left(x-5\right)+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x^2+1\right)\)
Biểu thức này không phân tích được thành nhân tử nữa bạn nhé.
x(x+1)^2+x(x-5)-5(x+1)^2
=(x+1)^2(x-5)+x(x-5)
=(x-5)((x+1)^2+x)
=(x-5)(x^2+3x+1)
~~~~~~~~~~~Ai đi qua nhớ để lại ~~~~~~~~~~~~~~~~~
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+1\right)\)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
Phân tích đa thức sau thành nhân tử:( x-5 )\(^2\)- 5 + x
Bài làm:
( x-5 )\(^2\)- 5 + x
= ( x - 5 )\(^2\)+ ( x - 5 ) Chỗ này có thể phân tích thành ( x - 5 ) ( x - 5 ) + ( x - 5 ) nha
= ( x - 5 ) \([\)( x - 5 ) + 1 \(]\)
= ( x- 5 ) ( x - 5 + 1 )
* Chúc bạn học tốt ^^
# Linh
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=y\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
x(x + 2) + x(x - 5) - 5(x + 2)
= [x(x + 2) - 5(x + 2)] + x(x - 5)
= (x - 5)(x + 2) + x(x - 5)
= (x - 5)(x + 2 + x)
= (x - 5)(2x + 2)
= 2(x - 5)(x + 1)