\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2014}{2015!}\)

\(=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)

\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)

\(=1-\frac{1}{2015!}< 1\left(đpcm\right)\)

1

1 đó

b)x³-2x²-4xy²+x

=x(x²-2x-4y²+1)

=x[(x²-2x+1)-4y²]

=x[(x-1)²-4y²]

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

c) (x+2)(x+3)(x+4)(x+5)-8

=[(x+2)(x+5)][(x+3)(x+4)]-8

=(x²+5x+2x+10)(x²+4x+3x+12)-8

=(x²+7x+10)(x²+7x+12)-8

đặt x²+7x+10 =a ta có

a(a+2)-8

=a²+2a-8

=a²+4a-2a-8

=(a²+4a)-(2a+8)

=a(a+4)-2(a+4)

=(a+4)(a-2)

thay a=x²+7x+10 ta đc

(x²+7x+10+4)(x²+7x+10-2)

=(x²+7x+14)(x²+7x+8)

bài 2 x³-x²y+3x-3y

=(x³-x²y)+(3x-3y)

=x²(x-y)+3(x-y)

=(x-y)(x²+3)

Đăng từng bài 1 thôi

1) \(\frac{x-y}{z-y}=-10\Leftrightarrow x-y=10\left(y-z\right)\)

\(\Leftrightarrow x-y=10y-10z\)

\(\Leftrightarrow x=11y-10z\)

Thay x=11y-10z vào biểu thức \(\frac{x-z}{y-z}\), ta có:

\(\frac{11y-10z-z}{y-z}=\frac{11y-11z}{y-z}=\frac{11\left(y-z\right)}{y-z}=11\)

Chá quá, có ghi nhìn không rõ đề

2) \(2x^2=9x-4\)

\(\Leftrightarrow2x^2-9x+4=0\)

\(\Leftrightarrow2x^2-8x-x+4=0\)

\(\Leftrightarrow2x\left(x-4\right)-1\left(x-4\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow2x-1=0\) hoặc x-4=0

1) 2x-1=0<=>x=1/2

2)x-4=0<=>x=4(Loại)

=> x=1/2

đáp án là : n(n+1)(n+2)

n(n+1)(n+2)