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a, (\(\frac{1}{2}\))x = \(\frac{1}{64}\)
<=> (\(\frac{1}{2}\))x = (\(\frac{1}{2}\))6
<=> x = 6
b, 35 :3x = 9
<=> 35 :3x = 32
<=> 3x = 35 :32
<=> 3x = 33
<=> x = 3
Chúc bạn học tốt
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Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=xy+x^2y^2+x^3y^3+...+x^{100}y^{100}\)
\(\Rightarrow A=xy+\left(xy\right)^2+\left(xy\right)^3+...+\left(xy\right)^{100}\)
\(\Rightarrow A=\left(-1\right)+1+\left(-1\right)+...+1\) ( 100 số hạng )
\(\Rightarrow A=\left[\left(-1\right)+1\right]+\left[\left(-1\right)+1\right]+...+\left[\left(-1\right)+1\right]\) ( 50 cặp số )
\(\Rightarrow A=0\)
Vậy A = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,a^2-2a+1-b^2\)
\(=\left(a^2-2a+1\right)-b^2\)
\(=\left(a-1\right)^2-b^2\)
\(=\left(a-1-b\right)\left(a-1+b\right)\) Khai triển thành hằng đẳng thức số 3 e nhé.
\(2,x^2+2xy+y^2-81\)
\(=\left(x^2+2xy+y^2\right)-81\)
\(=\left(x+y\right)^2-9^2\)
\(=\left(x+y-9\right)\left(x+y+9\right)\)Cái này cũng HĐT số 3 nè
\(3,x^2+6y-9-y^2\)
\(=-\left(y^2-6y+9\right)+x^2\)
\(=-\left(y-3\right)^2+x^2\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
\(5,4x^2+y^2-9-4xy\)
\(=\left(4x^2-4xy+y^2\right)-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y-3\right)\left(2x-y+3\right)\)
Học tốt
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{4}\cdot\frac{1}{16}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
b) \(81^x:9^x=729\)
\(\left(81:9\right)^x=729\)
\(9^x=9^3\)
\(\Rightarrow x=3\)
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{4}\times\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
Vậy x = 3
b) \(81^x:9^x=729\)
\(\Rightarrow\left(81:9\right)^x=729\)
\(\Rightarrow9^x=9^3\)
\(\Rightarrow x=3\)
Vậy x = 3
_Chúc bạn học tốt_