Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=\left(2x-1-x-3\right)\left(2x-1+x+3\right)=\left(x-4\right)\left(3x+2\right)\)
c) \(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
d) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-2^2\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
Phép tính b):
Đặt 2x - 1 = a ; x + 3 = b. Từ đầu bài suy ra:
\(\left(2x-1\right)^2-\left(x+3\right)^2\Rightarrow a^2-b^2\)
\(\Rightarrow a^2-b^2-\left(ab-ab\right)\Rightarrow\left(a^2-ab\right)-\left(b^2-ab\right)\)
\(\Rightarrow a\left(a-b\right)-b\left(b-a\right)\Rightarrow a\left(a-b\right)+b\left(a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)\)
Thế lại vào ta có:
\(\orbr{\begin{cases}a+b=\left(2x-1\right)+\left(x+3\right)=\left(2x+x\right)-\left(1-3\right)=3x+2\\a-b=\left(2x-1\right)-\left(x-3\right)=\left(2x-x\right)-\left(1-3\right)=x+2\end{cases}}\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)=\left(3x+2\right)\left(x+2\right)\)
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
1) \(x^2-2x-4y^2-4y\)
\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
2) \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^2\left(1-x^2\right)-4x+4x^2\)
\(=x^2\left(1+x\right)\left(1-x\right)+4x\left(x-1\right)\)
\(=x^2\left(1+x\right)\left(1-x\right)-4x\left(1-x\right)\)
\(=\left(1-x\right)\left[x^2\left(1+x\right)-4x\right]\)
1) \(25-x^2-y^2+2xy=5^2-\left(x^2-2xy+y^2\right)=5^2-\left(x-y\right)^2\)\(=\left(5-x+y\right)\left(5+x-y\right)\)
2) \(3x-3y-x^2+2xy-y^2\)\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)\(=3\left(x-y\right)-\left(x-y\right)^2\)\(=\left(x-y\right)\left(3-x+y\right)\)
1) \(25-x^2-y^2+2xy\)
\(=5^2-\left(x^2+y^2-2xy\right)\)
\(=5^2-\left(x-y\right)^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
2) \(3x-3y-x^2+2xy-y^2\)
\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)
\(=\left(3-x+y\right)\left(x-y\right)\)
a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
1/ \(3x^2+6x+3-3y^2=3x^2+3x+3x+3-3y^2\)
\(=3\left(x^2+2x+1-y^2\right)\)
\(=3\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=3\left[\left(x+1\right)^2-y^2\right]\)
\(=3\left(x+1-y\right)\left(x+1+y\right)\)
2/ \(25-x^2-y^2+2xy=5^2-\left(x^2+y^2-2xy\right)\)
\(=5^2-\left(x-y\right)^2\)
\(=\left[5-\left(x-y\right)\right]\left(5+x+y\right)\)
\(=\left(5-x+y\right)\left(5+x+y\right)\)
3/ \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3-\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3-x+y\right)\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left(\left(x-1\right)^2-y^2\right)\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
bài này bn hoc qt đung rui, tui làm nhé
1)= x(x2 -2x +1 - y) = x((x-1)2 - y)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
