a: A=-2^3*1/4+2*2^2*1/32-1/2*2*1/2

=-2+1/4-1/2

=-9/4

b: \(B=y^4\left(y^8+x^5y-100x^4\right)+100y^2\left(x^3-x^2\right)+100xy-6\)

=0+0+0-6=-6

c: \(C=-\dfrac{1}{2}\cdot\left(-\sqrt{3}\right)^2+25\cdot\dfrac{-1}{2}\cdot2-\sqrt{3}\cdot\dfrac{-1}{2}\cdot\left(-\sqrt{3}\right)\cdot8\)

=-3/2-25-4

=-3/2-29

=-30,5

câu a đó là sao ạ, t ko hiểu

(\(x\) - 2)² ≥ 0 \(\forall\) \(x\) 

-(\(x\) - 2)² ≤ 0 \(\forall\) \(x\)

Không có x nào thỏa điều kiện \(-\left(x-2\right)^2\ge0\)

a: Sửa đề: -y^2

x^2-4x+4-y^2

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

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

b: =(x-y)^2-(z-t)^2

=(x-y-z+t)(x-y+z-t)

\(45.\)

\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

\(=a^2-ab+b^2+3ab-6a^2b^2+6a^2b^2\)

\(=a^2+2ab+b^2\)

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

\(=1^2\)

\(=1\).

42:

a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b)-3bac

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

=0

=>a^3+b^3+c^3=3abc

44:

a: x^3+y^3+3xy

=(x+y)^3-3xy(x+y)+3xy

=1^3-3xy+3xy=1

b: x^3-y^3-3xy

=(x-y)^3+3xy(x-y)-3xy

=1^3+3xy-3xy=1

(a-b+c)^3

=[a+(-b)+c]^3

=

có phải ra như thế này ko mn :

a^3 - b^3 + c^3

a: 2x>0

=>x>0/2=0

b: -5x<=0

=>x>=0

c: -x<=0

=>x>=0

\(3.\)

\(a,\)

\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow4x^2-12x+9-x^2-10x-25=0\)

\(\Leftrightarrow3x^2-22x-16=0\)

\(\Leftrightarrow3.\left(x-8\right)\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3=0\left(\text{vô lí}\right)\\x-8=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(S=\left\{8;-\dfrac{2}{3}\right\}\)

\(b,\)

\(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{1;2\right\}\)

\(4.\)

\(a,\)

\(16x^3y+\dfrac{1}{4}yz^3\)

\(=\dfrac{1}{4}y\left(64x^3+z^3\right)\)

\(=\dfrac{1}{4}y\left(4x+z\right)\left(16x^2-4xz+z^2\right)\)

\(b,\)

\(x^{m+4}-x^{m+3}-x-1\)

\(=x^m.x^4-x^m.x^3-x-1\)

\(=x^m.\left(x^4-x^3\right)-x-1\)

\(=x^m.x^3.\left(x+1\right)-\left(x+1\right)\)

\(=\left(x^{m+3}-1\right)\left(x+1\right)\)

3:

a: =>(2x-3-x-5)(2x-3+x+5)=0

=>(x-8)(3x+2)=0

=>x=8 hoặc x=-2/3

b: =>x^3-x^2-4(x-1)^2=0

=>x^2(x-1)-4(x-1)^2=0

=>(x-1)(x^2-4x+4)=0

=>x=1 hoặc x=2

\(1.\)

\(a,\)

\(3x^2-6xy+3y^2\)

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

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

\(b,\)

\(12x^5y+24x^4y^2+12x^3y^3\)

\(=12x^3y\left(x^2+2xy+y^2\right)\)

\(=12x^3y\left(x+y\right)^2\)

\(c,\)

\(64xy-96x^2y+48x^3y-8x^4y\)

\(=8xy\left(8-12x+6x^2-x^3\right)\)

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

\(d,\)

\(54x^3+16y^3\)

\(=2\left(27x^3+8y^3\right)\)

\(=2\left[\left(3x\right)^3+\left(2y\right)^3\right]\)

\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)

\(2.\)

\(a,\)

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

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

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

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

\(b,\)

\(-16x^2+8xy-y^2+49\)

\(=49-\left(16x^2-8xy+y^2\right)\)

\(=7^2-\left(4x-y\right)^2\)

\(=\left(7-4x+y\right)\left(7+4x-y\right)\)

\(3.\)

\(a,\)

\(x^6-x^4+2x^3+2x^2\)

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

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

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

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

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

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

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

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

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

\(b,\)

\(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

1:

a: =3(x^2-2xy+y^2)

=3(x-y)^2

b: \(=12x^3y\left(x^2+2xy+y^2\right)=12x^3y\left(x+y\right)^2\)

c: \(=8xy\left(8-12x+6x^2-x^3\right)\)

=8xy(2-x)^3

d: =2(27x^3+8y^3)

=2(3x+2y)(9x^2-6xy+4y^2)

ĐK: \(\left\{{}\begin{matrix}y\ne1\\y\ne3\end{matrix}\right.\)

Biểu thức trở thành: \(\dfrac{\left(y+5\right)\left(y-3\right)}{\left(y-1\right)\left(y-3\right)}-\dfrac{\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=-\dfrac{8}{\left(y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow\dfrac{y^2-3y+5y-15}{\left(y-1\right)\left(y-3\right)}-\dfrac{y^2-1}{\left(y-1\right)\left(y-3\right)}+\dfrac{8}{\left(y-1\right)\left(y-3\right)}=0\)

\(\Leftrightarrow y^2+2y-15-y^2+1+8=0\\ \Leftrightarrow2y=6\\ \Leftrightarrow y=\dfrac{6}{2}=3\left(loại\right)\)

Vậy không có giá trị y để hai biểu thức trên bằng nhau.

\(\dfrac{y+5}{y-1}-\dfrac{y+1}{y-3}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)

=>(y+5)(y-3)-y^2+1=-8

=>y^2+2y-15-y^2+1=-8

=>2y-14=-8

=>2y=6

=>y=3(loại)

=(x^2+x)^2+3(x^2+x)+2-12

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

=(x^2+x+5)(x^2+x-2)

=(x+2)*(x-1)(x^2+x+5)