
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


x2 - 23x + 180 = 0
x(x - 23) = -180 (vô lí)
Vậy không có giá trị của x thõa mãn

ĐKXĐ: x<>5; x<>-6
\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>23x+61=2x+61
=>x=0




ta có x2+5x+4
=x2+x+4x+4
=(x2+x)+(4x+4)
=x(x+1)+4(x+1)
=(x+1)(x+4)
tương tự ta đc
x2+11x+28=(x+4)(x+7)
x2+17x+70=(x+7)(x+10)
x2+23x+130=(x+10)(x+13)
=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)
=> 13(x+13)-13(x+1)=4(x+1)(x+13)
=> 13[(x+13)-(x+1)]=(4x+4)(x+13)
=>13(x+13-x-1)=4x2+52x+4x+52
=13.12=4x2+56x+52
=>4x2+56x+52=156
=>4x2+56x-104=0


a) x2(5x3 – x - 1212) = x2. 5x3 + x2 . (-x) + x2 . (-1212)
= 5x5 – x3 – 1212x2
b) (3xy – x2 + y) 2323x2y = 2323x2y . 3xy + 2323x2y . (- x2) + 2323x2y . y
= 2x3y2 – 2323x4y + 2323x2y2
c) (4x3– 5xy + 2x)(- 1212xy) = - 1212xy . 4x3 + (- 1212xy) . (-5xy) + (- 1212xy) . 2x
= -2x4y + 5252x2y2 - x2y.
\(23x-x^2=120\)
\(\Leftrightarrow-x^2+23x-120=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15\\x=8\end{matrix}\right.\)
Vậy \(S=\left\{15;8\right\}\)
chúc mừng lên rank=)