Question

b1 phân tích đa thức thành nhân tử

a) x²-y²+10x+25

b) x³-x²-5x +125

c) x⁴+4y⁴

d)a(b²-c²)-b(c²-a²)+c(a²-b²)

e)7x²-10xy+3y²

f)a²+b³+c³-3abc

h) xy(x+y)-yz(y+z)+xz(x-z)

b2 tìm x,y

a)2x²-4-(3x+5)(5x+2)=0

b)4x-1=4x²

c)4x(x-3)-7(x-3)²

mấy bạn giải giúp mk vs mk đang cần gấp lắm

Answer 1

Bài 1:

a) \(x^2-y^2+10x+25\)

\(=\left(x^2+10x+25\right)-y^2\)

\(=\left(x+5\right)^2-y^2\)

\(=\left(x+y+5\right)\left(x-y+5\right)\)

b) \(x^3-x^2-5x+125\)

\(=x^3+5x^2-6x^2-30x+25x+125\)

\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

c) \(x^4+4y^4\)

\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)

\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)

d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

e) \(7x^2-10xy+3y^2\)

\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)

\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)

f) Sửa đề \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)

\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)

\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)

\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)

\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)

\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)

\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)

\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

Answer 2

Bài 2:

a: \(\Leftrightarrow2x^2-4-15x^2-6x-25x-10=0\)

=>-13x^2-31x-14=0

hay \(x\in\left\{\dfrac{-31+\sqrt{233}}{26};\dfrac{-31-\sqrt{233}}{26}\right\}\)

b: \(\Leftrightarrow4x^2-4x+1=0\)

=>(2x-1)^2=0

=>2x-1=0

=>x=1/2

c: Đề thiếu vế phải rồi bạn