đề bài chắc là trục căn thức ở mẫu nhỉ.
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}+1\right)^1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}+1}{x-1}\)
Cho A= \(\dfrac{1}{\sqrt{x}+10}\); B=\(\dfrac{4}{2-\sqrt{x}}\)
Tìm x để A nguyên
A = \(\dfrac{2\sqrt{x+1}}{\sqrt{x-3}}\)
Tìm x:
a) \(\dfrac{16}{3}:x=\dfrac{50}{12}\left(-0,06\right)\) b) \(2.x-1^3=-8\) c) \(\dfrac{\left(0.8\right)^5}{\left(0.4\right)^6}=x\)
d)\(0,944-2.x=3,268\) e)\(\sqrt{\dfrac{0,09}{25}=-\dfrac{4,7}{5}+x}\) g)\(\sqrt{5^2-3^2}=-\sqrt{81-x}\)
Cho biểu thức A=\(\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
a, Tính giá trị của A khi x=\(\dfrac{1}{4}\)
b, Tính giá trị của x để A=1
c,Tìm giá trị nguyên của x để A nhận giá trị nguyên
a,\(\sqrt{x+1}=7\)
b,(x-2).\(\left(x+\dfrac{2}{3}\right)\)>0
c,\(\left(\dfrac{2}{3}x-1\right).\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)
Bài 1:
a) \(\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)+\sqrt{2}.\dfrac{\sqrt{2^5}}{1-\sqrt{9}}\)
b)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right).\dfrac{49}{50}\)
Cảm ơn trước nha
Tìm x, biết
a.\(\dfrac{2}{3}\)x-\(\dfrac{1}{2}\)x = (\(\dfrac{-7}{12}\)) . \(\dfrac{7}{5}\)
b.(\(\dfrac{1}{5}\) - \(\dfrac{3}{2}\)x)\(^2\) = \(\dfrac{9}{4}\)
c. (1,25 - \(\dfrac{4}{5}\)x)\(^3\) = -125
d. 2\(^x\) + 2\(^{x+4}\) = 544
e.\(\dfrac{3}{4}\)x - 4,5+ \(\dfrac{4}{5}\)x = 3
g. \(\left|3x-2\right|\) =\(\left|2x-3\right|\)
h.(x2 -4)\(\sqrt{x}\) = 0
i. (\(\dfrac{1}{3}\)x - 4 )2 = \(\dfrac{16}{25}\)
j.4x2 -1 = 0
k.2x2+0,82 =1
l.\(\dfrac{5}{12}\)\(\sqrt{x}\) - \(\dfrac{1}{6}\) = \(\dfrac{1}{3}\)
m.(\(\sqrt{x-1}\) + 5)(x-6\(\sqrt{x}\) = 0
1.Tìm x biết rằng:
\(\sqrt{\left(\dfrac{1}{3}+x\right)^2}:\left(x+\dfrac{3}{4}\right)=\dfrac{7}{9}\)
Tìm x:
a) \(\dfrac{x}{-9}=\dfrac{-40}{45}\)
b) \(\left|2x+\dfrac{1}{3}\right|+\dfrac{4}{9}=-5+\sqrt{\dfrac{16}{81}}\)
