a: \(16x^3+0,25yz^3\)

\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)

\(=0,25\left(64x^3+yz^3\right)\)

b: \(x^4-4x^3+4x^2\)

\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)

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

c: \(x^3+x^2y-xy^2-y^3\)

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

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

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

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

d: \(x^3+x^2+x+1\)

\(=x^2\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+1\right)\)

e: \(x^4-x^2+2x-1\)

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

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

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

f: \(2x^2-18\)

\(=2\cdot x^2-2\cdot9\)

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

g: \(x^2+8x+7\)

\(=x^2+x+7x+7\)

\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

h: \(x^4y^4+4\)

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

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

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

i: \(x^4+4y^4\)

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

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

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

k: \(x^2-2x-15\)

\(=x^2-5x+3x-15\)

\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

\(a/\)

\(4x-4y+x^2-2xy+y^2\)

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

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

\(=\left(x-y\right)\left(4+x-y\right)\)

\(b/\)

\(x^4-4x^3-8x^2+8x\)

\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)

\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)

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

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

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

\(d/\)

\(x^4-x^2+2x-1\)

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

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

\(e/\)(Xem lại đề)

\(x^4+x^3+x^2+2x+1\)

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

\(=x^3\left(x+1\right)+\left(x+1\right)^2\)

\(=\left(x+1\right)\left(x^3+x+1\right)\)

\(f/\)

\(x^3-4x^2+4x-1\)

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

\(=x\left(x-2\right)^2-1\)

\(=[\sqrt{x}\left(x-2\right)]^2-1\)

\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)

\(c/\)

\(x^3+x^2-4x-4\)

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

\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+3x+2\right)\)

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

\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

= (2-x)(2+x)(x+y)

d) y^2+y-x^2+x

= (y^2 - x^2) + (x+y)

= (y-x)(y+x)+ (x+y)

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

= (2x-y)(2x+y) - (2x+y)

= (2x -y -1)(2x+y)

f)9x^2-25y^2-6x+10y

= 

ko biết làm

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

đúng rồi pn. giúp mik đc bài nào cũng đc

Bài 4 :

a) \(x^3+x^2y-xy^2-y^3=x^2\left(x+y\right)-y^2\left(x+y\right)=\left(x^2-y^2\right)\left(x+y\right)=\left(x-y\right)\left(x+y\right)^2\)

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

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

d)

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

e) Trùng câu d

f) \(x^3-y^3-3x+3y=\left(x-y\right)\left(x^2-xy+y^2\right)-3\left(x-y\right)=\left(x-y\right)\left(x^2-xy+y^2-3\right)\)

Bài 5:

a) \(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\

\(\Leftrightarrow2x-3=6\)

\(\Leftrightarrow x=\frac{9}{2}\)

vậy........

c) \(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)

Vậy

d) \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy ........