a) \(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

câu b đâu

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)\)

\(=\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=\frac{\left(a+b\right)\left(ab+ac+bc+c^2\right)}{abc\left(a+b+c\right)}\)

\(=\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc\left(a+b+c\right)}\)

2.

pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0

<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0

<=> (x-2000).(1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004) = 0

<=> x-2000=0 ( vì 1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004 > 0 )

<=> x=2000

Tk mk nha

1.

a, = (2x-1)^2-2.(2x-1)+1-4

    = (2x-1-1)^2-4

    = (2x-2)^2-4

    = (2x-2-2).(2x-2+2)

    = 2x.(2x-4)

b, = [x.(x+3)].[(x+1).(x+2)]

    = (x^2+3x).(x^2+3x+1)-8

    = (x^2+3x+1)^2-1-8

    = (x^2+3x+1)^2-9

    = (x^2+3x+1-3).(x^2+3x+1+3)

    = (x^2+3x-2).(x^2+3x+4)

    = ((x+1).(x+3).(x^2+3x-2)

Tk mk nha

bạn có thể giải giùm  mk bài toán hình nx k ạ

\(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}=\left(x+\frac{1}{2}\right)^3\)

Bạn ghi sai đề nha

Hok tốt

\(x^3+\frac{3}{2}x^2+\frac{3}{2}x+\frac{1}{8}\)

\(=\left(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\right)+\frac{3}{2}x-\frac{3}{4}x\)

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

\(=\left(x+\frac{1}{2}\right)^3+\left(\sqrt[3]{\frac{3}{4}x}\right)^3\)

\(=\left(x+\frac{1}{2}+\sqrt[3]{\frac{3}{4}x}\right)\left[\left(x+\frac{1}{2}\right)^2-\left(x+\frac{1}{2}\right)\left(\sqrt[3]{\frac{3}{4}}\right)+\left(\sqrt[3]{\frac{3}{4}}\right)^2\right]\)

Cho mk xin ghi lại đề nha :

\(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

Nhớ zúp >.<

Mấy bài này mình đã làm rồi. 

a.(x+1)(x+2)(x+3)(x+4)-24=[(x+1)(x+4)][(x+2)(x+3)]-24=(\(x^2+5x+4\))(\(x^2+5x+6\))-24  (1)

đặt \(x^2+5x+5=a\)ta có (1)=(a-1)(a+1)-24=\(a^2-25=\left(a-5\right)\left(a+5\right)\)

thay a=\(x^2+5x+5\)vào (1) ta có (1)=(\(x^2+5x\)+5-5)(\(x^2+5x\)+5+5)=x(x+5)(\(x^2\)+5x+10)

b.ta có :\(\frac{a}{3}+\frac{a^2}{2}+\frac{a^3}{6}=\frac{2a+3a^2+a^3}{6}=\frac{a\left(a^2+3a+2\right)}{6}\)=\(\frac{a\left(a^2+2a+a+2\right)}{6}=\frac{a\left(a+1\right)\left(a+2\right)}{6}\).ta lại có a(a+1)(a+2) là tích 3 số nguyên liên tiếp luôn chia hết cho 6 suy ta điều cần cm

\(\frac{2}{3}x-\frac{1}{9}x^2-1\)

\(=-\left(\frac{1}{9}x^2-\frac{2}{3}x+1\right)\)

\(=-\left[\left(\frac{1}{3}x\right)^2-2\cdot\frac{1}{3}x\cdot1+1^2\right]\)

\(=-\left(\frac{1}{3}x-1\right)^2\)

\(x^2-\frac{5}{3}x-\frac{2}{3}\)

\(=x^2-2x+\frac{1}{3}x-\frac{2}{3}\)

\(=x\left(x-2\right)+\frac{1}{3}\left(x-2\right)\)

\(=\left(x-2\right)\left(x+\frac{1}{3}\right)\)