Question

[nguyen thi thuy linh]

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) \(8^x:2^x=4\)

Answer 1

[Phương Trâm]

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 -.-

Answer 2

[Nịna Hatori]

- 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