tim x , biet :
a) x + \(\dfrac{3}{10}\)= \(\dfrac{-2}{5}\)
b) x + \(\dfrac{5}{6}\) = \(\dfrac{2}{5}\) - \(\left(-\dfrac{2}{3}\right)\)
c) \(1\dfrac{2}{5}\) x + \(\dfrac{3}{7}\) = \(-\dfrac{4}{5}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
e) \(\left[x+\dfrac{4}{15}\right]-\left[-3,75\right]=-\left[-2,15\right]\)
f) \(\left(x-2\right)^2=1\)
g) \(\left(2x-1\right)^3=-27\)
h) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
i) x : \(\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
j)...
Đọc tiếp
tim x , biet :
a) x + \(\dfrac{3}{10}\)= \(\dfrac{-2}{5}\)
b) x + \(\dfrac{5}{6}\) = \(\dfrac{2}{5}\) - \(\left(-\dfrac{2}{3}\right)\)
c) \(1\dfrac{2}{5}\) x + \(\dfrac{3}{7}\) = \(-\dfrac{4}{5}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
e) \(\left[x+\dfrac{4}{15}\right]-\left[-3,75\right]=-\left[-2,15\right]\)
f) \(\left(x-2\right)^2=1\)
g) \(\left(2x-1\right)^3=-27\)
h) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
i) x : \(\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
j) \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
k) \(8^x:2^x=4\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1