\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{2019a^2}{2019c^2}=\frac{2020b^2}{2020d^2}=\)

\(=\frac{2019a^2+2020b^2}{2019c^2+2020d^2}=\frac{2019a^2-2020b^2}{2019c^2-2020d^2}\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)

Bạn ơi tham khảo thử cách này nhé !

Từ  \(\frac{a}{b}=\frac{c}{d}\)( bài cho )

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó :

+) \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019\left(bk\right)^2+2020b^2}{2019\left(bk\right)^2-2020b^2}=\frac{b^2\left(2019k^2+2020\right)}{b^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)

+) \(\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019\left(dk\right)^2+2020d^2}{2019\left(dk\right)^2-2020d^2}=\frac{d^2\left(2019k^2+2020\right)}{d^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)

ghi đề kĩ dùm cái bạn ơi

Gọi số học sinh lần lượt ở khối 7,8,9 là x,y,z.

Ta có : \(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :\(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}=\frac{b-a}{6-5}=\frac{50}{1}=50\)

\(=>\frac{a}{5}=50=>a=250\)

\(=>\frac{b}{6}=50=>b=300\)

\(=>\frac{c}{7}=50=>c=350\)

Vậy :.....................

\(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)

\(=>-\frac{1}{3}< x-\frac{3}{5}< \frac{1}{3}\)

Vì x thuộc Z nên \(x-\frac{3}{5}=0\)

Vậy : \(x=\frac{3}{5}\)

\(1.\frac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x\div\left(-3\right)^4=\left(-3\right)^3\)

\(\Rightarrow\left(-3\right)^x=\left(-3\right)^7\Rightarrow x=7\)

\(2.\sqrt{x-5}-4=5\Rightarrow\sqrt{x-5}=9\Rightarrow\sqrt{x-5}=\sqrt{81}\Rightarrow x-5=81\Rightarrow x=86\)

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