Số hạng đó là số hạng thứ 4 \(\Rightarrow k=3\) nên có dạng:

\(C_6^3\left(2x\right)^3.\left(-y^2\right)^3=-C_6^3\left(2x\right)^3y^6\)

`a)A` có nghĩa `<=>x-1 >= 0 <=>x >= 1`

`b)B=\sqrt{3^2 .2}+\sqrt{2^3}-\sqrt{5^2 .2}`

`<=>B=3\sqrt{2}+2\sqrt{2}-5\sqrt{2}`

`<=>B=0`

`c)` Với `a >= 0,a \ne 1` có:

`C=[a-1]/[\sqrt{a}-1]-[a\sqrt{a}-1]/[a-1]`

`C=[(a-1)(\sqrt{a}+1)-a\sqrt{a}+1]/[(\sqrt{a}-1)(\sqrt{a}+1)]`

`C=[a\sqrt{a}+a-\sqrt{a}-1-a\sqrt{a}+1]/[(\sqrt{a}-1)(\sqrt{a}+1)]`

`C=a/[a-1]`

14A

15B

16B

17C

18B

19B

20C

\(16.\\ \left\{{}\begin{matrix}P+N+E=76\\P=E\\\left(P+E\right)-N=20\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P+N=76\\2P-N=20\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P=E=Z=24\\N=28\end{matrix}\right.\)

\(20.\\ \left\{{}\begin{matrix}P+E+Z=40\\\left(P+E\right)-N=12\\P=E\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P+N=40\\2P-N=12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P=E=Z=13\\N=14\end{matrix}\right.\\ \Rightarrow A=Z+N=13+14=27\left(đ.v.C\right)\)

\(19\\ \left\{{}\begin{matrix}P+E+N=40\\P=E\\\dfrac{N}{P+E}=\dfrac{7}{13}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P+N=40\\\dfrac{N}{2P}=\dfrac{7}{13}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P+N=40\\13N-14P=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P=E=Z=13\\N=14\end{matrix}\right.\\ \Rightarrow A=Z+N=13+14=27\left(đ.v.C\right)\\ \Rightarrow KH:^{27}_{13}R\)

=> Chọn B

Câu 58: B

Câu 59: C

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)