\(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Leftrightarrow\left(2x-2\right)-\left(6x+6\right)-\left(8x+12\right)-16=0\)
\(\Leftrightarrow2x-2-6x-6-8x-12-16=0\)
\(\Leftrightarrow-12x-36=0\)
\(\Leftrightarrow-12x=36\)
\(\Leftrightarrow x=\dfrac{36}{-12}=-3\)
Vậy x=-3
Tìm x biết : a) x+3/22=-27/121.11/9 b) (3và2/20-2và1/20)x+2/7=3/5 c) |3/2x-5|-1/2=-1/4
Cho các đa thức :
F(x)=x^3.(3x-1)-x(1+3x^4)
G(x)=x^2(x^2+2)-x(x^4+2x^2+7)+3
H(x)=x^3(-2+2x-x^2)-1/2(5x-3-2x^2)
a) Tính F(x)+G(x)-H(x)=A(x)
F(x)-G(x)-H(x)=B(x)
F(x)+G(x)-2H(x)=C(x)
b) Tìm nghiệm của C(x)
tìm x
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) - 0
c, x - 2\(\sqrt{x}\) = 0
d, (x - 1)\(^2\) + 1/7 = 0
1. tìm x biết
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) = 0
c, 5 - x\(^2\) = 1
d, x - 2\(\sqrt{x}\) = 0
g, (x - 1) + \(\dfrac{1}{7}\) = 0
giúp mik vs ạ
bài 1
a tìm số thực x biết: |x - 25/3| -2=7/3
b cho hàm số y=ʃ(x)= 3-\(^{2x^2}\) . tính ʃ(1/2)
Tìm x,y,z biết :
1) \(\dfrac{x}{5}=\dfrac{y}{4}\)và x2 - y2=4
2) \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
3) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)và 2x + 3y - z = 95
a)1/2x+3/5x=-33/22
b) (2/3x -4/7).(1/2+-3/7:x)=0
c) (4/5-2x).(1/3+3/5:x)=0
1/ Tìm x biết:
a. (2x-1)3= -27
b, (2x-3)4= 625
c. (x-2)5=(x-2)7
d. 5x+2 + 5x+3 =750
Tim x,y:
|x-4|+3|3x-2|-2|2-x|=x+1
||2x-3|-3/4|=5/6
|1/2-|3-x||=2x-1
|x-2007|+2004|y-2008|\(\le\)0
3|x-2y|+4|y+1/2|=0
