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a)\(A=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\left(ĐK:x\ne0;-5\right)\)
\(\Leftrightarrow A=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}\)
\(\Leftrightarrow A=\frac{x+5}{5}\)
b)Để A=-4 \(\Leftrightarrow\frac{x+5}{5}=-4\)
\(\Leftrightarrow x+5=-20\)
\(\Leftrightarrow x=-25\)
a).....
\(=\frac{x^2}{5\left(x+5\right)}+\frac{2x-10}{x}+\frac{50+5x}{x\left(x+5\right)}\) MTC= 5x (x+5) ĐK\(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\frac{x^2.x}{5x\left(x+5\right)}+\frac{5.\left(2x-10\right).\left(x+5\right)}{5x\left(x+5\right)}+\frac{5.\left(50+5x\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+\left(10x-50\right).\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+50x-50x-250+250+25x}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
b) A=-4
=>\(\frac{x+5}{5}=-4\)
=> x = -25
c)
d) Để A đạt gt nguyên thì 5\(⋮\)x+5
=> \(\left(x+5\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
*x+5=1 => x=-4 \(\in Z\)
*x+5=-1 => x=-6\(\in Z\)
*x+5=5 => x=0\(\in Z\)
*x+5=-5 => x=-10\(\in Z\)
Vậy...........
a )
\(A=x\left(x^3+y\right)-x^2\left(x^2-y\right)-x^2\left(y-1\right)\)
\(\Rightarrow A=x^4+xy-x^4+x^2y-x^2y+x^2\)
\(\Rightarrow A=x^2+xy=x\left(x+y\right)\)
Thay \(x=-10;y=5\)vào A , ta được :
\(A=-10\left(-10+5\right)\)
\(=-10.-5=50\)
Vậy \(A=50\)
a) A = x(x3 + y) - x2(x2 - y) - x2(y - 1)
= x4 + xy - x4 + x2y - x2y + x2
= xy + x2
Thay x = –10 và y = 5 vào (1), ta được:
A = -10.5 + (-10)2 = -50 + 100 = 50
Vậy giá trị của biểu thức A tại x = –10 và y = 5 là 50.
b)Ta có: 5x3 – 3x2 + 10x – 6 = (5x3 + 10x )+ ( -3x2– 6)
= 5x(x2 + 2) – 3(x2 + 2) = (x2 + 2)(5x – 3)
Vậy (x2 + 2)(5x – 3) = 0 ⇒ 5x – 3 = 0 (vì x2 + 2 ≥ 0, với mọi x)
⇒x = 3/5
c)Ta có: x2 + y2 – 2x + 4y + 5 = (x2 – 2x + 1) + (y2 + 4y + 4)
= (x – 1)2 + (y + 2)2
Vậy (x – 1)2 + (y + 2)2 = 0 ⇒ x – 1 = 0 hay y + 2 = 0
⇒ x = 1 hoặc y = -2
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
b) B= 4 .(x-6) - x2.(2+3x)+ x .(5x-4)+ 3x2 .(x-1)
=4x-24-2x2-3x3+5x2-4x+3x3-3x2
=-3x3+3x3-2x2+5x2-3x2+4x-4x-24
=-24
vậy giá trị của B ko phụ thuộc vào biến x
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
a: \(A=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x^2-15x}-\dfrac{8}{5x^2+10x}\right):\dfrac{x^2-2x+2}{x^2-x-6}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x\left(x-3\right)}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4}{5x}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x+8-8}{5x\left(x+2\right)}\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x}{5x}\cdot\dfrac{x-3}{x^2-2x+2}=\dfrac{\left(5x+4\right)\left(x-3\right)}{5\left(x^2-2x+2\right)}\)
b: Khi x=1 thì \(A=\dfrac{\left(5+4\right)\left(1-3\right)}{5\left(1-2+2\right)}=\dfrac{9\cdot\left(-2\right)}{5}=\dfrac{-18}{5}\)
Khi x=3 thì \(A=\dfrac{\left(5\cdot3+4\right)\left(3-3\right)}{A}=0\)