Tim x biet (x-2)(3-x)< 0
a, Cho F(x) = a x+b . Tim a,b biet f(0) = 3 va F(2) =-1
b, Cho F(x) =a x+ b. Tim a,b biet F(1) = -1 va F(-2) = 8
c, Cho F(x) =a x +b .tim a,b biet F(0) = 1 va F(-2) = -9
Tim x biet:
(x-3)^2-x/x-3/=0
(x-3)2 - x/ x-3/ =0
+ Nếu x>/3
=> (x-3)2 -x (x-3) =0
=> (x-3)(x-3-x) =0 => -3(x-3) =0 => x-3=0 => x =3 (TM)
+Nếu x<3
=> (x-3)2 +x(x-3) =0
=> (x-3)(x-3+x) =0 => (x-3)(2x-3) =0 => x =3 ( loại) ; 2x-3 =0 => x =3/2 (TM)
Vậy x thuộc {3;3/2}
tim x,biet:
a)x(x-2)+x-22=0
b)5x(x-3)-x+3=0
\(5x\left(x-3\right)-x+3=0\)
<=> \(\left(x-3\right)\left(5x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
Vay..........
Tim x biet:3*x+x2=0
3x + X^2 = 0
X( 3+X) = 0
=> X= 0 HOẶC 3+X = 0
=> X=0 HOẶC X= -3
Tim x biet : 20 . 2^x + 1 = 10.4^2 + 1
Tim x : ( 4-x:2)^3 - 1 = 2 . (2^3 - 5 : 2^0 )
20 . 2^x + 1 = 10.4^2 + 1
20 . 2^x + 1 = 10 . 16 + 1
20 . 2^x + 1 = 161
20 . 2^x = 161 - 1
20 . 2^x = 160
2^x = 8
2^x = 2^3
=> x = 3
( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )
( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )
( 4 - x : 2 )^3 - 1 = 2 . 3
( 4 - x : 2 )^3 - 1 = 6
( 4 - x : 2 )^3 = 7
=> ko tìm đc x
tim x thuoc Z biet x^3-x^2+x-1=0
tim x biet : (X+1/2)x(X-3/4)=0
\(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
=> x = \(\frac{-1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
=> x = \(\frac{3}{4}\)
tim x biet (x^2-1)(x^2-3)(x^2-5)(x^2-7)<=0
tim x thuộc q biet ( x - 2) nhân ( x+2/3 ) > 0