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Ta có: \(\frac{a}{b}=\frac{3}{4}\) => \(\frac{a}{3}=\frac{b}{4}\) . Đặt đẳng thức \(\frac{a}{3}=\frac{b}{4}=k\)
=> a = 3k ; b = 4k
=> \(a^2=9k^2\) ; \(b^2=16k^2\)
Lại có: \(A=\frac{a^2+b^2}{a^2-b^2}=\frac{9k^2+16k^2}{9k^2-16k^2}=\frac{25k^2}{-7k^2}=\frac{25}{-7}\)
Vậy A = \(-\frac{25}{7}\)
Chúc bạn học tốt !!
\(\frac{a}{b}=\frac{3}{4}\)
\(A=\frac{a^2+b^2}{a^2-b^b}=\frac{3^2+4^2}{3^2-4^4}=-\frac{25}{247}\)
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\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{2}\)
\(\Rightarrow\dfrac{x^3}{125}=\dfrac{y^3}{64}=\dfrac{z^3}{8}=\dfrac{x^3-y^3+z^3}{125-64+8}=\dfrac{69}{69}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt[3]{125}=5\\y=\sqrt[3]{64}=4\\z=\sqrt[3]{8}=2\end{matrix}\right.\)
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Ta có : \(\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=4.3=12\)
k cho mk nha!
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Ta có :
x2 - y2 = 4
\(\frac{x}{5}=\frac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x}{5}=\frac{y}{3}=\left(\frac{x}{5}\right)^2=\left(\frac{y}{3}\right)^2=\frac{x^2}{5^2}=\frac{y^2}{3^2}=\frac{x^2-y^2}{5^2-3^2}=\frac{4}{16}=\frac{1}{4}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{4}.5=\frac{5}{4}\\y=\frac{1}{4}.3=\frac{3}{4}\end{cases}}\)
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(\(\frac{1}{5}\))2 .n = (\(\frac{1}{125}\))3 - n
<=> \(\frac{1}{25}\)n +n = \(\frac{1}{5^9}\)
<=> \(\frac{26}{25}\)n = \(\frac{1}{5^9}\)
<=> n = \(\frac{1}{5^9}\): \(\frac{26}{25}\)= \(\frac{1}{2031250}\)
\(\left(\frac{2}{3}\right)^3=\frac{8}{27}\)
\(\frac{2}{3^3}=\frac{2}{27}\)
\(\left(\frac{2}{3}\right)^3=\left(\frac{2^3}{3^3}\right)=\frac{8}{27}\)