a) \(\left(x+y\right)^3-x^3-y^3\)

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

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

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

\(=3xy\left(x+y\right)\)

b) \(x^2+y^2+2xy+yz+xz\)

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

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

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

c) \(x^2-10xy-1+25y^2\)

\(=\left(x^2-10xy+25y^2\right)-1\)

\(=\left(x-5y\right)^2-1\)

\(=\left(x-5y-1\right)\left(x-5y+1\right)\)

d) \(ax^2-ax+bx^2-bx+a+b\)

\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)

\(=x^2(a+b)-x(a+b)+(a+b)\)

\(=(a+b)(x^2-x+1)\)

e)\(x^2-2y+3xz+x-2y+3z\)

\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)

\(=x(x+1)-2y(x-1)+3z(x+1)\)

\(=(x+1)(x-2y+3z)\)

f) \(xyz-xy-yz-xz+x+y+z-1\)

\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)

\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)

\(=(z-1)(xy-y-x+1)\)

\(=(z-1)[y(x-1)-(x-1)]\)

\(=(z-1)(x-1)(y-1)\)

_Học tốt_

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

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

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

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

d) \(3xy-z-3x+yz=3x\left(y-1\right)-x\left(y-1\right)=2x\left(y-1\right)\)

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

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

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

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

c, \(x^2-2x+1-y^2+2y-1=\left(x-1\right)^2-\left(y-1\right)^2\)để thế này đc thôi 

d, \(3xy-z-3x+yz=2x\left(y-1\right)\)

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

                        A B C D

a) Vì ABCD là hình thang ( \(AB//CD\))

\(\Rightarrow\widehat{B}+\widehat{C}=180^o\)

mà \(\widehat{B}-\widehat{C}=50^o\)\(\Rightarrow\widehat{B}=\frac{180^o+50^o}{2}=115^o\)

\(\Rightarrow\widehat{C}=180^o-115^o=65^o\)

Vì \(AB//CD\)\(\Rightarrow\widehat{A}+\widehat{D}=180^o\)

mà \(\widehat{A}=\frac{1}{3}.\widehat{D}\)\(\Rightarrow\frac{1}{3}.\widehat{D}+\widehat{D}=180^o\)

\(\Rightarrow\frac{4}{3}.\widehat{D}=180^o\)\(\Rightarrow\widehat{D}=135^o\)\(\Rightarrow\widehat{A}=\frac{1}{3}.135^o=45^o\)

Vậy \(\widehat{A}=45^o\); \(\widehat{B}=115^o\); \(\widehat{C}=65^o\); \(\widehat{D}=135^o\)

Vì ABCD là hình thang ( AB // CD )

\(\Rightarrow\hept{\begin{cases}\widehat{A}+\widehat{B}=180^o\\\widehat{C}+\widehat{D}=180^o\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\widehat{B}=\left(180+50\right):2=165^o\\\widehat{C}=165-50=95^o\end{cases}}\)

+) \(\widehat{A}=\frac{1}{3}\widehat{D}\)

\(\Rightarrow\widehat{D}=3\widehat{A}\)

\(\Rightarrow\widehat{A}+\widehat{D}=\widehat{A}+3\widehat{A}=4\widehat{A}=180^o\)

\(\Rightarrow\widehat{A}=180:4=45^o\)

\(\widehat{D}=3\widehat{A}=45.3=135^o\)

Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .

a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

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

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

c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)

d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)

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

\(\Leftrightarrow2xy-3x-2+y^2=0\)

\(2x\left(8x-1\right)^2\left(4x-1\right)-9=512x^4-256x^3+40x^2-2x-9\)

\(=\left(2x-1\right)\left(4x+1\right)\left(64x^2-16x+9\right)\)

\(x^2+2xy+y^2=x^2+3x+2\)

\(\Leftrightarrow x^2-x^2+2xy+3x+y^2-2=0\)

\(\Leftrightarrow2xy+3x+y^2-2=0\)

P/s : chả hiểu đề bài :))

Vì AB // CD nên \(\hept{\begin{cases}\widehat{A}+\widehat{D}=180^0\\\widehat{B}+\widehat{C}=180^0\end{cases}}\)(định lí hình thang)

Mà \(\widehat{A}=5\widehat{D}\)=> \(\widehat{5D}+\widehat{D}=180^0\)=> \(6\widehat{D}=180^0\)=> \(\widehat{D}=30^0\)(1)

Thay (1) vào \(\widehat{A}=5\widehat{D}\)ta có :

\(\widehat{A}=5\cdot30^0=150^0\)

Lại có : \(\widehat{B}=4\widehat{C}\)

=> \(4\widehat{C}+\widehat{C}=180^0\)

=> \(5\widehat{C}=180^0\)

=> \(\widehat{C}=36^0\)(2)

Thay (2) vào \(\widehat{B}=4\widehat{C}\)ta có :

=> \(\widehat{B}=4\cdot36^0=144^0\)

Vậy : ^A = 150⁰ , ^B = 144⁰ , ^C = 36⁰ , ^D = 30⁰

a) 2( x - 1 )² + ( x + 3 )² = 3( x - 2 )( x + 1 )

<=> 2( x² - 2x + 1 ) + x² + 6x + 9 = 3( x² - x - 2 )

<=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

<=> 2x² - 4x + x² + 6x - 3x² + 3x = -6 - 2 - 9

<=> 5x = -17

<=> x = -17/5

b) ( x - 1 )² - 2( x - 3 ) = ( x + 1 )²

<=> x² - 2x + 1 - 2x + 6 = x² + 2x + 1

<=> x² - 2x - 2x - x² - 2x = 1 - 1 - 6

<=> -6x = -6

<=> x = 1 

c) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )² + 3x² = -33

<=> x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 ) + 3x² = -33

<=> x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6 + 3x² = -33

<=> x³ - 9x² + 27x - x³ + 6x² + 12x + 3x² = -33 - 27 + 27 - 6

<=> 39x = -39

<=> x = -1

a) Đặt \(a=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x+3=a+4\\x-2=a-1\\x+1=a+2\end{cases}}\)

    Ta có: \(2a^2+\left(a+4\right)^2=3.\left(a-1\right)\left(a+2\right)\)

        \(\Leftrightarrow2a^2+a^2+4a+4=3.\left(a^2+a-2\right)\)

        \(\Leftrightarrow3a^2+4a+4=3a^2+3a-6\)

        \(\Leftrightarrow a=-10\)

         \(\Rightarrow x-1=-10\)

        \(\Leftrightarrow x=-9\)

Vậy \(S=\left\{-9\right\}\)

b)  Đặt \(b=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x-3=b-2\\x+1=b+2\end{cases}}\)

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

         \(\Leftrightarrow b^2-2b+4=b^2+4b+4\)

         \(\Leftrightarrow-6b=0\) 

         \(\Leftrightarrow b=0\)

          \(\Rightarrow x-1=0\)

         \(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

c) Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

        \(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x^2+2x+1\right)+3x^2+33=0\)

        \(\Leftrightarrow6x^2+12x+6+3x^2+33=0\)

       \(\Leftrightarrow9x^2+12x+39=0\)

       \(\Leftrightarrow\left(9x^2+12x+4\right)+35=0\)

       \(\Leftrightarrow\left(3x+2\right)^2+35=0\)

   Vì \(\left(3x+2\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(3x+2\right)^2+35\ge35>0\forall x\)

         mà \(\left(3x+2\right)^2+35=0\)

   \(\Rightarrow\)\(\left(3x+2\right)^2+35=0\)vô nghiệm

Vậy \(S=\varnothing\)