a)  -1/8 -2x/5-1/3=3

-2x/5=3+1/8+1/3

-2x/5=83/24

-2x=(83×5)/24=415/24

x = (415÷-2)/24= -415/48

b) -7/3 -(25/6 -4/3+ 3/2)

= -7/3 -13/3 = -20/3

ai ma co 10 k ban chac ban phai lap chuc nink thui

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

k nha

Nhân 2A lên rồi lấy 2A-A là ra kết quả

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(\Rightarrow A=2A-A\)

\(\Rightarrow A=2-\frac{1}{2^{2012}}\)

a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)

\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)

\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)

\(x=-\frac{1}{3}\)

Vậy...

c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

=>\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

a) Ta có: \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)

\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3-8x^3+\frac{1}{27}\)

\(=8x^3+\frac{1}{27}-8x^3+\frac{1}{27}\)

\(=\frac{2}{27}\)

Vậy: Giá trị của biểu thức \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\) không phụ thuộc vào biến

b) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)

\(=x^3-3x^2+3x-1-\left(x^3-1\right)-3x\left(1-x\right)\)

\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)

\(=0\)

Vậy: Giá trị của biểu thức \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\) không phụ thuộc vào biến

c) Ta có: \(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)

\(=yx^4-y^5-yx^4+y^5\)

\(=0\)

Vậy: Giá trị của biểu thức \(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\) không phụ thuộc vào biến

Tìm x:

a)-2³+0,5x=1,5

-8+0,5x = 1,5

0,5x = 1,5-(-8) = 9,5

x = 9,5:0,5 = 19

b)(−3)x81=−27

(-3)^x:81 =-27

(-3)^x = -27.81 = -2187

(-3)^x = (-3)⁷

=> x=7

c)112.x−4=0,5

1,5.x = 0,5+4 = 4,5

x = 4,5:1,5 = 3

d)123:x4=6:0,3

\(\frac{5}{3}\):\(\frac{x}{4}\) = 20

\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)

=> x:4 = \(\frac{1}{12}\)

x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)

a) \(x^3-\frac{4}{25}x=0\)

\(\Leftrightarrow x\left(x+\frac{2}{5}\right)\left(x-\frac{2}{5}\right)=0\)

<=> x = 0

Xét 2 trường hợp: 

\(\Leftrightarrow x+\frac{2}{5}=0\)

      \(x=0-\frac{2}{5}\)

      \(x=-\frac{2}{5}\)

\(\Leftrightarrow x-\frac{2}{5}=0\)

      \(x=0+\frac{2}{5}\)

      \(x=\frac{2}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{2}{5}\end{cases}}\)

b) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{4}{3}\)

\(=\frac{13}{40}:\frac{4}{3}\)

\(=\frac{39}{120}=\frac{13}{40}\)

c) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)

\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)

\(=-\frac{5}{2}-1\)

\(=-\frac{7}{2}\)

\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)

\(=-\left(2x+\frac{7}{5}\right)^m\)

đến đây thì mình chịu