giúp mình với

\(x^2.\left(x-4\right)-x+4=0\)

\(\Rightarrow x^2.\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right).\left(x^2-1\right)=0\)

\(\Rightarrow\left(x-4\right).\left(x-1\right).\left(x+1\right)=0\)

Trường hợp 1: \(x-4=0\Rightarrow x=4\)

Trường hợp 2: \(x-1=0\Rightarrow x=1\)

Trường hợp 3: \(x+1=0\Rightarrow x=-1\)

\(A=x^2+4x+8\)

\(=x^2+4x+4+4\)

\(=\left(x^2+4x+4\right)+4\)

\(=\left(x+2\right)^2+4\)

Mà: \(\left(x+2\right)^2\ge0\forall x\Rightarrow\left(x+2\right)^2+4\ge4\forall x\)

Dấu '' = '' xảy ra khi: \(x+2=0\Rightarrow x=0\)

Vậy giá trị nhỏ nhất của \(A=4\) khi \(x=2\)

a) \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)

b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)

d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)

e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)

nnnvnvvnvnvnnvneech

\(ax-bx+ay-by-3a+3b\)

\(=x\left(a-b\right)+y\left(a-b\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(x+y-3\right)\)

x^4+2x^3-13x^2-14x+24

=x^4-x^3+3x^3-3x^2-10x^2+10x-24x+24

=(x^4-x^3)+(3x^3-3x^2)-10x^2+10x-24x+24

=x^3.(x-1)+3x^2(x-1)+10x(x-1)-24(x-1)

=(x-1).(x^3+3x^2+10x-24)

=(x-1).(x^3-3x^2+6x^2-18x+8x-24)

=(x-1).[(x^3-3x^2)+6x^2-18x+(8x-24)]

=(x-1).[x^2.(x-3)+6x(x-3)+8(x-3)]

=(x-1).(x-3).(x^2+6x+8)

=(x-1).(x-3).(x^2=4x+2x+8)

=(x-1).(x-3).[x(x+4)+2(x+4)]

=(x-1).(x-3).(x+2)(x+4)

Bài III:

2) (Xem trong ảnh)

Bài 3: 

\(\frac{x-5}{x-2}-\frac{x+4}{2x-x^2}\left(ĐK:x\ne2;x\ne0\right)\)

\(=\frac{x-5}{x-2}+\frac{x+4}{x^2-2x}\)

\(=\frac{x.(x-5)}{x.\left(x-2\right)}+\frac{x+4}{x.\left(x-2\right)}\)

\(=\frac{x^2-5x+x+4}{x.\left(x-2\right)}\)

\(=\frac{x^2-4x+4}{x.\left(x-2\right)}\)

\(=\frac{\left(x-2\right)^2}{x.\left(x-2\right)}\)

\(=\frac{x-2}{x}\)

\(\frac{x-3}{x+2}+\frac{4x}{x-3}-\frac{8x+4x^2}{x^2-x-6}\)

\(=\frac{\left(x-3\right)^2+4x.\left(x+2\right)-8x-4x^2}{\left(x-3\right).\left(x+2\right)}\)

\(=\frac{x^2-6x+9+4x^2+8x-8x-4x^2}{\left(x-3\right).\left(x+2\right)}\)

\(=\frac{\left(x-3\right)^2}{\left(x+2\right).\left(x-3\right)}\)

\(=\frac{x-3}{x+2}\)

a) ⇒x2+8−x3+x+3x=2⇒x2+8−x3+x+3x=2

⇒4x=−6⇒x=−32⇒4x=−6⇒x=−32

b) ⇒[x=02x2=3⇒[x=02x2=3⇒⎡⎣x=0x2=32⇒[x=0x2=32⇒⎡⎢⎣x=0x=±√32

b) 9x^2 - 1 = (3x + 1)*(4x + 1)
<=> (3x + 1)(3x - 1) - (3x + 1)(4x + 1) = 0
<=> (3x + 1)[3x - 1 - (4x + 1)] = 0
<=> (3x + 1)(3x - 1 - 4x - 1) = 0
<=> (3x + 1)(-x - 2) = 0
<=> 3x + 1 = 0
<=> x = -1/3
hoặc -x - 2 = 0
<=> x = -2
vậy S = { -2 ; -1/3 }

a) Theo đề ra: \(\hept{\begin{cases}a+b=p\left(1\right)\\a-b=q\left(2\right)\end{cases}}\)

Trừ các vế tương ứng (1) và (2)

\(\left(a+b\right)-\left(a-b\right)=p-q\)

\(\Rightarrow a+b-a+b=p-q\)

\(\Rightarrow b=\frac{p-q}{2}\) \(\left(11\right)\)

Cộng các vế tương ứng (1) và (2)

\(\left(a+b\right)+\left(a-b\right)=p+q\)

\(\Rightarrow a+b+a-b=p-q\)

\(\Rightarrow a=\frac{p+q}{2}\left(22\right)\)

Từ (11) và (22) ta có:

\(a.b=\frac{p+q}{2}.\frac{p-q}{2}\)

\(\Rightarrow ab=\frac{p^2-q^2}{4}\)

b) Ta có: \(a^2+b^2=\left(a^2+2ab+b^2-2ab\right)=\left(a+b\right)^2-2ab\)

Mà đề ra: \(a+b=p\) và theo phần a) ta có \(ab=\frac{p^2-q^2}{4}\)

\(a^2+b^2=p^2-2.\frac{p^2-q^2}{4}\Rightarrow a^2-b^2=\frac{p^2-q^2}{2}\)

\(a^3+b^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)

Mà đề ra: \(a+b=p\) và \(ab=\frac{p^2-q^2}{4}\) và \(a^2+b^2=\frac{p^2-q^2}{2}\)

\(a^3+b^3=p.\left(\frac{p^2+q^2}{2}-\frac{p^2-q^2}{4}\right)\Rightarrow a^3+b^3=p.\frac{p^2+3q^2}{4}\Rightarrow a^3+b^3=\frac{p^3+3pq^2}{4}\)