(2x - 3)2 - (x + 5)2 = 0
=> (2x - 3 - x - 5).(2x - 3 + x + 5) = 0
=> (x - 8).(3x + 2) = 0
=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)=> \(\orbr{\begin{cases}x=8\\3x=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=8\\x=\frac{-2}{3}\end{cases}}\)
Vậy \(x\in\left\{8;\frac{-2}{3}\right\}\)
Tim x , biet
(2x-1) ^2 + (x+3)^2 -5(x+7)(x-7) = 0
moi ng lam nhanh giup ! Thanks
Tim x, biet :
(x+2)(x^2-2x+4)+x(5-x)(x+5)=0
Tim x, biet:
(2x-1)2-(2x+3).(2x-3)=0
Tim x biet
\(x^4-2x^3-2x^2+3x+2=0\)
tim x biet:
x+x2-x3-x4=0
2x3+3x2+2x2+3=0
x2-x-12=0
Tim x biet: x+\(2\sqrt{2x^2}\) +2x3=0
tim x biet
(2x=1)^2 - 4(x=2)^2=9
3(x-1)^2 -3x(x-5)=1
3(x+2)^2+ (2x-1)^2 =7
7(x+3)(x-3)=36
tim x biet x2-11x+18=0
-4x2+5x-1=0 2x3+3x2+2x+3=0x(2x-7)-4x+14=0tinh nhanh bt sau
A=3.(x-3).(x+7)+(x-4)2+48,tai x=0.5
tim x biet
a) 9x^2+6x+1=0
b) 25x^2=4
c) 8-125x^3=0
d) (2x+1)-10(2x+1)(x+2)+25(x+2)^2=0
GIÚP MIK VS NHÉ
