a: \(\Leftrightarrow2x=0-\dfrac{3-\sqrt{13}}{2}=\dfrac{\sqrt{13}-3}{2}\)
hay \(x=\dfrac{\sqrt{13}-3}{4}\)
b: \(\Leftrightarrow2x+\dfrac{3+\sqrt{13}}{2}=0\)
\(\Leftrightarrow2x=\dfrac{\sqrt{13}+3}{2}\)
hay \(x=\dfrac{\sqrt{13}+3}{4}\)
a: \(\Leftrightarrow2x=0-\dfrac{3-\sqrt{13}}{2}=\dfrac{\sqrt{13}-3}{2}\)
hay \(x=\dfrac{\sqrt{13}-3}{4}\)
b: \(\Leftrightarrow2x+\dfrac{3+\sqrt{13}}{2}=0\)
\(\Leftrightarrow2x=\dfrac{\sqrt{13}+3}{2}\)
hay \(x=\dfrac{\sqrt{13}+3}{4}\)
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)
a) - x - \(\dfrac{2}{3}=-\dfrac{6}{7}\) b) (5x - 1). (2x - \(\dfrac{1}{3}=0\)
tìm gtln hoặc gtnn: a,p= 3.|2x-y|+2\(\sqrt{x-3}\) b,n=-4\(\sqrt{6}\) -3x-7 c,h=4(x-2y)\(^8\) = 2\(\sqrt{y+2+3}\) d,s= ,\(\dfrac{-7}{3\sqrt{ }x-4+2\left(x-3y^{ }\right)^{ }4+2}+3\)
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)\(\begin{matrix}&\left(\dfrac{2}{3}\right)^0\end{matrix}-\sqrt{9+}\left(-\dfrac{^{ }1}{2}\right)^2\)
Bài 1:Tìm GTLN của các bểu thức:
a) A=5-3.(2x-1)2
b) B=\(\dfrac{1}{2.\left(x-1\right)^2+3}\)
c) C=\(\dfrac{x^2+8}{x^2+2},x\in Z\)
Bài 2: Tìm x sao cho:
a)(x-1)(x-2)>0
b) (x-2)2.(x+1).(x-4)<0
c) \(\dfrac{5}{x}< x\)
Làm hộ mình nhé
HElP ME!!
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, \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
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
\(4x^2-1=0\)
\(2x^2_{ }+0.82=1\)
\(\left(3x-\dfrac{1}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)
\(\dfrac{2x+1}{2}=\dfrac{3x-1}{3}=\dfrac{y-2}{4}\)