\(\frac{\left(-3\right)^x}{81}=-27\)
\(1\frac{1}{2}.x-4=0,5\)
2x-1=16
lam on giai day du ho minh nka minh se tich cho
a,\(\left(\frac{-1}{2}\right)^3-\left(\frac{2}{5}x+\frac{1}{3}\right)=3\)
b,\(-2\frac{1}{3}-\left(4\frac{1}{6}-\frac{4}{3}+1\frac{1}{2}\right)\)
AI GIAI DUOC MINH CHO 10 TICK.NHANH NHA MINH DANG CAN GAP
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
cho x,y>0 và 2x>y Chứng minh rằng \(\left(\frac{1}{x}+2\right)^2.\left(\frac{2}{y}-\frac{1}{x}\right).\frac{2y-1}{y}< =\frac{81}{8}\)
Tìm x biết :
4,\(\frac{\left(-3\right)^x}{81}=-27\)
5,\(1\frac{1}{2}.x-4=0,5\)
6,\(2^{x-1}=16\)
7,\(\left(x-1\right)^2=25\)
8,\(|2x-1|=5\)
9,\(0,2-|4,2-2x|=0\)
11,\(1\frac{2}{3}:\frac{x}{4}=6:0,3\)
12,\(2\frac{2}{3}:x=1\frac{7}{9}:2\frac{2}{3}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2} +\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
ghi loi giai day du nha may ban
ai ghi du loi giai va nhanh minh se tick(toan lop 6 nhe)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
k nha
Nhân 2A lên rồi lấy 2A-A là ra kết quả
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow A=2A-A\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
Tìm x biết
a) \(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
b) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
d) \(\left(x+\frac{1}{4}\right)^4=\frac{16}{81}\)
Ai xong nhanh nhất cho 1 like \(☺\)
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
chứng minh rằng giá trị biểu thức sau ko hụ thuộc vào biến
a.\(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)
b.\(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)
c.\(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
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
b)\(\frac{\left(-3\right)^x}{81}=-27\)
c)\(1\frac{1}{2}.x-4=0,5\)
d)\(1\frac{2}{3}:\frac{x}{4}=6:0,3\)
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,\(x^3-\frac{4}{25}x=0\)
b,\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
c\(4.\left(\frac{-1}{2}\right)^3-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)-1.\left(\frac{-1}{2}\right)^0\)
AI GIAI DUOC MINH CHO 10 TICK
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) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)
\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)
\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)
\(=-\frac{5}{2}-1\)
\(=-\frac{7}{2}\)
\(\left[-\frac{2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(-\frac{2}{5}x^3\right)\)
tim x nguyen
giai ra giup minh voi
\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu