1)Tìm x, biết :
\(4.\left[3x-1\right]+\left[x\right]-2.\left[x-5\right]+7.\left[x-3\right]=12\)
\(\left[2\dfrac{1}{5}-x\right]+\left[x-\dfrac{1}{5}\right]+8\dfrac{1}{5}=1,2\)
\(3.\left[x+4\right]-\left[2.x+1\right]-5.\left[x-3\right]+\left[x-9\right]=5\)
\(2\left[x+3\dfrac{1}{2}\right]+\left[x\right]-3\dfrac{1}{2}=\left[2\dfrac{1}{5}-x\right]\)
Tìm x biết:
a) \(\left(-\dfrac{2}{3}\right)^2.x=\left(-\dfrac{2}{3}\right)^5\) ; b) \(\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\) ; c) (2x-3)\(^2\) ; d) (3x-2)\(^5\) =-243
\(\left(\dfrac{32}{243}\right)^x=\left(\dfrac{3}{2}\right)^{-1}\)
a, \(\dfrac{\left(-3\right)^x}{81}=-27\)
b, \(2^{x-1}=16\)
c, \(\left(x-1\right)^2=25\)
d, \(0,2-\left|4,2-2x\right|=0\)
e, \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
tính: a) 12.\(\left\{-\dfrac{2}{3}\right\}^2+\dfrac{4}{3}\) b) 12,5.\(\left\{\dfrac{-5}{7}\right\}+1,5.\left\{\dfrac{-5}{7}\right\}\)
Câu 1:Thực hiện phép tính(tính một cách hợp lí nếu có thể):
a)\(\dfrac{1}{2}-\dfrac{-3}{6}=+\dfrac{5}{3}-\dfrac{9}{12}\)
b)\(\begin{matrix}&\left(\dfrac{-2}{3}\right)\end{matrix}.\dfrac{3}{11}+\left(\dfrac{-16}{9}\right):\dfrac{11}{3}\)
c)\(\begin{matrix}&\left(\dfrac{2}{3}\right)^0\end{matrix}-\sqrt{9+}\left(-\dfrac{^{ }1}{2}\right)^2\)
\(4x^2-1=0\)
\(2x^2_{ }+0.82=1\)
\(\left(3x-\dfrac{1}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)
bài 1 tìm x thuộc Q biết
a. |x|=\(^1_53\) và x<0
b.|x|=-2,1
c.|x-3,5|=5
d. |x+\(\dfrac{3}{4}\)|-\(\dfrac{1}{2}\)=0
e. |x-\(\dfrac{2}{5}\)|+\(\dfrac{1}{2}\)=\(\dfrac{3}{4}\)
f. \(\dfrac{5}{6}\)-|2-x|=\(\dfrac{1}{3}\)
g. (2x-5)^2=9
h. \(\sqrt{3-7x}\)=\(\dfrac{1}{4}\)
i. (\(\dfrac{2}{3}\))^x=\(\dfrac{8}{27}\)
k. (x+\(\dfrac{1}{2}\))^2=\(\dfrac{1}{9}\)
l. \(\dfrac{\left(-3^x\right)}{81}\)=-27
m.\(\left(x-2\right)\)^2x+3=(x-2)^2x+1(x thuộc N)
Câu 2: Tìm x,y,z biết:
a) \(\dfrac{x+y}{2014}=\dfrac{xy}{2015}=\dfrac{x-y}{2016}\)
b) \(\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|=2016\)
Xin lỗi vì đăng ko đug dạng bài nhưng mk mong các bn giúp mk vs ak. mk cảm ơn
tìm x:
(x-3)+2x2-6x=0
tính:
A=\(\left|x+\frac{3}{2}\right|-\left|x-1\right|+\left|x-\frac{3}{4}\right|\)
