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Minhf ko vieets ddc daaus banj owi!

help me!

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Một thùng và bốn can đựng đc số l xăng là:

         47+ 4✖9=83 (l)

Banj lamf 2 pheps tinhs dduwowcj ko aj?

a^2+b^2/ 9+16=102/25

=>a^2=109/25.9=981/25=>a=căn 981/25

=>b^2=109/25.16=1744/25=>b= căn 1744/25

a²/9 = b²/6 = (a²+b²)/(9+6) = 102/15 = 6,8 (tính chất dãy tỉ số bằng nhau)

=> a² = 6,8.9 = 61,2 = > a = \(\sqrt{61,2}\)

b² = 6,8.16 = 108,8 => b = \(\sqrt{108,8}\)

immoral

immoral

immoral nhé bạn

\(\text{≌₰⇴⩸⨙⩸※◡⨦}\)

1.
\(A=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{\left(x^2-4\right)+x-2}{x^2-4}=\dfrac{\left(x-2\right)\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(x+2+1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
thay x = 98 ta được: \(A=\dfrac{101}{100}\)
2. (đkxd \(x\ne\pm1\))
\(B=\dfrac{x-1}{x+1}+\dfrac{x+1}{x-1}+\dfrac{5x}{1-x^2}=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{5x}{\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2+\left(x+1\right)^2-5x}{x^2-1}=\dfrac{x^2-2x+1+x^2+2x+1-5x}{x^2-1}=\dfrac{2x^2-5x+2}{x^2-1}=\dfrac{2x^2-4x-x+2}{x^2-1}=\dfrac{2x\left(x-2\right)-\left(x-2\right)}{x^2-1}=\dfrac{\left(x-2\right)\left(2x-1\right)}{x^2-1}\)để B bằng 0 thì: \(\left(x-2\right)\left(2x-1\right)=0\left(x^2-1\ge0\forall x\ne\pm1\right)\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)