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Bài 1: Tính
\(\left(\frac{-2}{3}\right)^4\times9^2\)
Bài 2: Tìm x
\(x^4=\frac{16}{625}\)
Giups mk vs!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
1. \(\left(-\frac{2}{3}\right)^4.9^2\)
\(=\frac{16}{81}.81\)
\(=16\)
2. \(x^4=\frac{16}{625}\)
=>\(x^4=\left(\frac{2}{5}\right)^4\)
=> \(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
# Học tốt #
Bài 1:
\(\left(-\frac{2}{3}\right)^4.9^2=\frac{\left(-2\right)^4}{3^4}.\left(3^2\right)^2=\frac{2^4}{3^4}.3^4=2^4=16\)
Bài 2:
\(x^4=\frac{16}{625}\Leftrightarrow x^4=\frac{4^4}{5^4}\Leftrightarrow x^4=\left(\frac{4}{5}\right)^4\)
\(\Rightarrow x=\frac{4}{5}\)hoặc\(x=-\frac{4}{5}\)
Vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
5^(x-2)(x+3) = 1 <=> 5^(x-2)(x+3) = 5x^0
<=> (x-2)(x+3) = 0
<=> x=2 hoac x=-3
Vậy x\(\in\left\{2;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) 3x + 4 + 3x + 2 = 810
=> 3x . 34 + 3x . 32 = 810
=> 3x.(34 + 32) = 810
=> 3x . (81 + 9) = 810
=> 3x . 90 = 810
=> 3x = 810 : 90
=> 3x = 9
=> 3x = 32
=> x = 2
d) 3x + 3x + 2 = 810
=> 3x + 3x . 32 = 810
=> 3x . (1 + 32) = 810
=> 3x . (1 + 9) = 810
=> 3x . 10 = 810
=> 3x = 810 : 10
=> 3x = 81
=> 3x = 34
=> x = 4
c, \(3^{x+4}+3^{x+2}=810\)
\(\Leftrightarrow3^x\left(3^4+3^2\right)=810\)
\(\Leftrightarrow3^x.90=810\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
d, \(3^x+3^{x+2}=810\)
\(\Leftrightarrow3^x\left(1+3^2\right)=810\)
\(\Leftrightarrow3^x.10=810\)
\(\Leftrightarrow3^x=81=3^4\)
\(\Leftrightarrow x=4\)
P/s: Toán thường thôi nhỉ :) Ko nâng cao lắm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|\)
\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|x-1-x-5\right|\)
\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|-6\right|=6\)
dấu "=" xảy ra khi \(\left(x-1\right).\left(x+5\right)\ge0\)
\(\Rightarrow-5\le x\le1\)
Vậy x={-5,-4,-3,-2,-1,0,1}
b) \(\hept{\begin{cases}\left(2x-y+3\right)^4\ge0\\\left|y+2\right|\ge0\end{cases}}\)
mà \(\left(2x-y+3\right)^4+\left|y+2\right|=0\)
dấu "=" xảy ra khi \(\hept{\begin{cases}\left(2x-y+3\right)^4=0\\\left|y+2\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=-2\end{cases}}\)
vậy \(x=-\frac{5}{2},y=-2\)
dấu "=" xảy ra khi
Vậy x={-5,-4,-3,-2,-1,0,1}
b)
mà
dấu "=" xảy ra khi
vậy
Phần a hình như sai hay sao ý
\(b,3^x+3^{x+2}=810\)
\(\Rightarrow3^x+3^x\cdot9=810\)
\(\Rightarrow3^x\cdot\left(1+9\right)=810\)
\(\Rightarrow3^x\cdot10=810\)
\(\Rightarrow3^x=810:10=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
a)2^4-2^x=16^4
2^4-2^x=2^16
2^x=2^4-2^16
2^x=2^-12
x=-12
câu b mik ko biet nha ban !!:))