\(\left|x\right|< 2\Rightarrow\left[{}\begin{matrix}x< 2\\x>-2\end{matrix}\right.\)
\(=>-2< x< 2\)
nghiệm của bpt được biểu diễn trên trục số
giải bpt
\(x\in\left\{-2;2\right\}\)
4.Tìm x ,biết
a,2.(x-2)-3=x+2
b,3-2.(3-x)=3.(x-10)-17
c,2./x-2/-3=15
d,9x+3)^2=16
e,(5-x)^3=-27
g,x.(x-2)=(x-1).(x-2)
h,3./x-2/+2./x-2/=15
k,(x-2).(x-3)=(x-3).(x-4)
m,(x+1)^2=(x+2)^2
-2(x mũ 2 - 3x+2)+ 3(x mũ 2 - 4)
4(x-x mũ 2+2) + 3(x mũ 2 - 2x-4)
-5(x+2) - 3(x+1)
-2(x-3) - x(x-3)
-6(x+2)- (x+1)3
\(5^{x+4}-ba.5^{x+ba}=2.5^{11}\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+ba}+2^{x+4}+2^{x+5}=480\)
1 rút gọn biểu thức
A=(x-1)(x+2)-(x-2)(x-3)-x(x-3)+x(x+2)
B=(2x-3)(x-2)+(x+1)(2-x)-(x-2)(x-1)
|x+3|+|-x-3|=20
-(-1-x)+(-x)^3+x(x^2-2)=-30
2*x*x*x+3*(-x)^2*x+4*(-x)^3-5*(-x)=-10
x^2(x-2)-16x+32=0
|x+3|=|2x-9|
(2x+9)^2=(x+15)^2
|x-15|-|2x+6|=0
Tìm x,biết:\(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2021}=2^{2025}+8\)
a,2x(x+5)=(x+3)2+(x-1)2+20
b,(2x-2)2=(x+1)2+3(x-2)(x+5)
c,(x-1)2+(x+3)2=2(x-2)(x+1)+38
d,(x+2)3-(x-2)3=12x(x-1)-8
D = 3 { x + 1 } + 2 { x - 2 } - x + 3
E = 7 { 2 x + 1 } - 3 { x - 2 } + 5
F = { x - 2 } { x + 2 } - { x - 3 } { x + 3 } -1
G = { x - 1 } { x + 1 } - x { x - 3 } + 2
a,(x^5)^10=x
b,2^x.3^x=4
c,4^x+2=64
d,2^x.2^2^2=2^3^2
e,5^2x+2=25^x+3
g,(x-2)^2=9.16
h,(x-1)^2=(x-10)^4
a,1/3 .(x-2/5)=3/4 b, 7/3:(x-2/3)=4/5 c,1/3.(x-2/5)=4/5 d, 2/3.(x-1/2)-1/4.(x-2/5)=7/3 e,3/7 .(x-2/3)+1/2=5/4.(x-2) f,1/2.(x-3)+1/3.(x-4)+1/4.(x-5)=1/5 g,[2/3.(x-1/2)-4/5]:(x-1/3)=21/5 h, {x-[1/2.(x-3)+11/5]}:(x-1/2)=3/5 i,x.(x-2/5)-(x+2).x+11/4=4/3
