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\(a.\left(\frac{2}{5}\right)^5:\left(\frac{9}{25}\right)^5=\left(\frac{2\cdot25}{9\cdot5}\right)^5=\frac{10}{9}^5\)
\(b.25\cdot5^3\cdot\frac{1}{625}\cdot5^2=\frac{5^7}{5^4}=5^3\)
\(c.\frac{20^5\cdot5^{10}}{100^5}=\frac{2^{10}\cdot5^{15}}{2^{10}\cdot5^{10}}=5^5\)
\(d.\frac{1}{7}^2\cdot\frac{1}{7}\cdot49^2=\frac{7^4}{7^3}=7\)
Ta có: F= (100-12) (100-22)...(100-252)
=> F= (100-12)...(100-102)...(100-252)
=> F= (100-12)...0...(100-252)
=> F= 0
Vậy F= 0
\(\left|x-1\right|-\left(-2^3\right)=9.\left(-1\right)^{100}\)
\(\left|x-1\right|+8=9.1\)
\(\left|x-1\right|+8=9\)
\(\left|x-1\right|=9-8\)
\(\left|x-1\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
\(\frac{x-2}{-4}=\frac{-9}{x-2}\)
\(\Rightarrow\left(x-2\right)\left(x-2\right)=\left(-4\right).\left(-9\right)\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^2=6^2\\\left(x-2\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)
\(\frac{x-5}{3}=\frac{-12}{5-x}\)
\(\Rightarrow\left(x-5\right).\left(5-x\right)=\left(-12\right).3\)
\(\Rightarrow5x-25-x^2-5x=-36\)
\(\Rightarrow25-x^2=-36\)
\(x^2=25-\left(-36\right)\)
\(\Rightarrow\orbr{\begin{cases}x^2=61\\x^2=-61\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{61}\\x=\sqrt{-61}\end{cases}}\)
a) \(\left|x-1\right|-\left(-2\right)^3=9\left(-1\right)^{100}\)
\(\left|x-1\right|+8=9\)
\(\left|x-1\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
vay \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)
\(\left(x-2\right)^2=-9.\left(-4\right)\)
\(\left(x-2\right)^2=36\)
\(\left(x-2\right)^2=6^2\)
\(\Rightarrow x-2=6\)
\(\Rightarrow x=8\)
vay \(x=8\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)
\(\frac{x-5}{3}=\frac{12}{x-5}\)
\(\left(x-5\right)^2=12.3\)
\(\left(x-5\right)^2=36\)
\(\left(x-5\right)^2=6^2\)
\(\Rightarrow x-5=6\)
\(\Rightarrow x=11\)
vay \(x=11\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
Ta có :5/x = 1/8 - y/4 = (1-2y)/8
<=> x = 5.8/(1-2y) ; thấy 1-2y là số lẻ nên ƯCLN(8,1-2y) = 1
do đó x/8 = 5/(1-2y)
Để x, y nguyên khi 1-2y phải là ước của 5
*Xét 1-2y = -1 => y = 1 => x = -40
*Xét 1-2y = 1 => y = 0 => x = 40
*Xét 1-2y = -5 => y = 3 => x = -8
*Xét 1-2y = 5 => y = -2 => x = 8
Vậy có 4 cặp (x,y) nguyên (-40,1) ; (40, 0) ; (-8, -5) ; (8, 5)
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian