bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

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

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3\left(x^2+2x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x^2+x+x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left[x\left(x+1\right)+\left(x+1\right)\right]}\)

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

nè bn thành cái chỗ X²+2x+1 đã là HĐT rồi thì tách làm j nữa dài đó

Ta thấy: \(1-\frac{2}{n.\left(n+1\right)}=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-1-1}{n.\left(n+1\right)}=\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)

Lại có: \(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right).....\left(1-\frac{2}{99.100}\right)\)

\(=\left(1-\frac{2}{2.\left(2+1\right)}\right).\left(1-\frac{2}{3.\left(3+1\right)}\right).\left(1-\frac{2}{4.\left(4+1\right)}\right).....\left(1-\frac{2}{99.\left(99+1\right)}\right)\)

\(=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}.\frac{\left(4-1\right).\left(4+2\right)}{4.\left(4+1\right)}.....\frac{\left(99-1\right).\left(99+2\right)}{99.\left(99+1\right)}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{98.101}{99.100}\)

\(=\frac{1.4.2.5.3.6.....98.101}{2.3.3.4.4.5.....99.100}\)

\(=\frac{\left(1.2.3.....98\right).\left(4.5.6.....101\right)}{\left(2.3.4.....99\right).\left(3.4.5.....100\right)}\)

\(=\frac{1.101}{99.3}\)

\(=\frac{101}{297}\)

Gọi ba số dó có dang x ; x + 2 ; x + 4 

THeo bài ra ta có :

( x+ 2 )( x+ 4 ) - x ( x+  2 ) = 192 

=> x^2 + 2x + 4x + 8 - x^2 - 2x = 192 

=> 4x + 8  = 192

=> 4x = 192 - 8 

=> 4x = 184 

=> x = 46 

Vậy ba số chẵn là 46 ; 48 ; 50 

Gọi ba số chẵn liên tiếp đó là:a;b;c. Ta có:

b.c-a.b=192

(c-a)b  =192

4b       =192

=>b     =48

Vậy a=46; c=50

Ta có :

( x- 5 )( 2x + 3 ) -  2x ( x- 3 ) + x + 7 

= 2x^2 + 3x - 10x - 15 - (2x^2 - 6x ) +x + 7 

= 2x^2 - 7x - 15 - 2x^2 + 6x + x + 7 

= ( 2x^2 - 2x^2 ) + ( -7x + 6x + x ) + 7 - 15 

= 0 + 0 + ( -8) = -8 

Vậy GT biểu thức không phụ thuộc vào biến