40 phut = 2/3 gio

Quang duong di ve dai :

  30 x 2/3 = 20 ( km )

Quang duong luc di : 

  20 - 5 = 15 ( km )

KL : ...

mk lớp 6

Dễ mà bạn 

X + y = 5 => x = 5 - y

=> \(\left(5-y\right)^2+y^2=53\)

=>\(25-10y+2y^2=53\)

Đễn đây bạn tìm đc y nhờ giải pt bậc 2 và để tìm x bạn thay y vào " x = 5-y"

\(x^2+y^2=53\)

\(\Leftrightarrow\left(x+y\right)^2-2xy=53\)

\(\Leftrightarrow-2xy=28\)

\(\Leftrightarrow xy=-14\)

Do đó ta có các cặp (x;y) là: (-1;14); (14;-1); (-14;1); (-1;14); (-7;2); (2;-7); (7;-2);(-2;7)

( 12x - 5 )( 4x - 1 ) + ( 3x - 7  )( 1 - 16x ) = 81

( 12x.4x ) + ( 12x.1) + ( -5.4x) + ( -5).(-1) + 3x.1 + 3x. (-16x) + (-7).1 + ( -7 ).(-16x) = 81

\(48^2-12x-20x+5\)\(+3x-48x^2-7+112x\)\(=81\)

\(\left(-12x-20x\right)\)\(+\left(5-7\right)\)\(+\left(3x+112x\right)\)\(=81\)

\(-32x-2+115x^2\)\(=81\)

\(\left(-32x^2+115x^2\right)-2\)\(=81\)

\(83x-2\)\(=81\)

\(83x=81+2\)

\(83x=83\)

\(\Rightarrow x=1\)

 ( 12x - 5 )( 4x - 1 ) + ( 3x - 7 )( 1 - 16x ) = 81

<=> 48x²-32x+5 +3x-48x²-7+112x=81

<=> 83x= 81 +2

<=> 83x=83

<=> x=1

= \(x^2-xy+x-y^2-y\)\(+xy\)

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

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

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

no need to :)

Ta có: \(a^2+b^2+c^2+d^2=a\left(b+c+d\right)\)

   \(\Leftrightarrow4\left(a^2+b^2+c^2+d^2\right)=4a\left(b+c+d\right)\)

   \(\Leftrightarrow4a^2+4b^2+4c^2+4d^2=4ab+4ac+4ad\)

   \(\Leftrightarrow a^2-4ab+4b^2+a^2-4ac+4c^2+a^2-4ad+4d^2+a^2=0\)

   \(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+a^2=0\)

   \(\Leftrightarrow a-2b=0,a-2c=0,a-2d=0,a=0\)[4 ptrinh này bạn để trong dấu''{" ].

    \(\Leftrightarrow a=b=c=d=0\)

M = ( x - 5)( x + 2 ) + ( 3x - 6 )( x + 2 ) - ( 3x - 1/2  )² + 5x² 

= x²-3x-10+3x²-12-(9x²-3x+1/4)+5x²

= x²-3x-10+3x²-12-9x²+3x-1/4+5x²

= 0.x - 89/4

Thay x=2018 => M= -89/4

\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)

\(M=x^2+2x-5x-10+\left(3x^2+6x-6x-12\right)-\left(9x^2-\frac{3}{2}x+\frac{1}{4}\right)+5x^2\)

\(M=x^2-3x-10+3x^2-12-9x^2+\frac{3}{2}x-\frac{1}{4}+5x^2\)

\(M=-\frac{3}{2}x-\frac{41}{4}\)

Thay x = 2018 vào biểu thức \(M=-\frac{3}{2}x-\frac{41}{4}\), ta có:

\(M=-\frac{3}{2}.2018-\frac{41}{4}=-3027-\frac{41}{4}=\frac{-12149}{4}\)

Vậy giá trị của biểu thức \(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)khi x = 2018 là \(-\frac{12149}{4}\)

me too

a) x(x-5)+2(x-5) = (x-5)(x+2)
b) (x-7)(x-2)

c) (x+2)(x^2+2x+4)+5y(x+2) = (x+2)(x^2+2x+4+5y) 

d) (x^2+8)^2 -16x^2 = (x^2+8-4x)(x^2+8+4x)

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

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

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

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

\(\left(a-b\right)^2\)

\(=\left(a-b\right)\left(a-b\right)\)

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

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

Thanks