\(\left[{}\begin{matrix}x+4=\sqrt{6}\\x+4=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4+\sqrt{6}\\x=-4-\sqrt{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+4=\sqrt{6}\\x+4=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4+\sqrt{6}\\x=-4-\sqrt{6}\end{matrix}\right.\)
Tìm x, biết:
a, (x+8).(x+6)-x^2=104
b, (x+1).(x+2)-(x-3).(x+4)=6
c, 3.(2x-1).(x+2)-2.(3x+2).(x-4)=5
tìm x
1}x+3[x-1]=4 2}2[x-3]+5=3
3}x[x-2]-x^2=-2 4}x^2-x[x+2]=6 5}3x[x-5]-3x[x-3]=6
6}3[x^2-2x+1]+x[2-3x]=7
tìm x biết (x-2)2-(x-3)(x+3)=6
(x-4)2-(x-2)(x+2)=6
a, 5*(4x-1)+2*(1-3x)-6*(x+5)=10
b, 2x*(x+1)+3*(x-1)*(x+1)-5x*(x+1)+6x mũ 2 = 0
c, 4*(x-1)*(x+5)-(x+2)*(x+5)-3(x-1)*(x+2)=0
d,2*(5x-8)-3*(4x-5)=4*(3x-4)+11
\(\dfrac{1}{2x-x^2-4}\) tìm GTLN/ GTNN
\(\dfrac{3x^2+14}{x^2+4}\)
\(\dfrac{2x+1}{x^2+2}\)
Tìm x biết : x.(2x2 + x -2 ) - 2x3 - x. (x + 3) - 1 = 4
tìm x:
a) ( x + 5) ( x-1) - ( x-5) ( x-2) =0
b) ( x-3) (x+7) - ( x-2) ( x+4) = 0
c) ( x- 6) (x-7) - x( x+4) -5(x-1) = 0
1) Cho x+y=a, x2+y2=b, x3+y3=c
C/m a3-3ab+2c=0
2)Cho x2+y2=1
Tính 2(x6+y6)-3(x4+y4)
giải phương trinh
a) (x^2 - 1)( x^2 + 4 )= 0
b) x-3/5 = 6
tìm x
a)4(2x+1)2+(4x+2)(2-6x)+(3x-1)2=0
b) (6-x+4x2)(6+x-4x2)=36