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Đề là ntn:
\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)
Đề là Thực hiện biến đổi toán học và kết hợp với máy tính . Tính số nghịch đảo của biểu thức ?
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
15
\(\dfrac{7}{x-2}\)+\(\dfrac{8}{x-5}\)=3 (x khác 2 khác 5)
\(\Leftrightarrow\)7*(x-5)+8(x-2)=3(x-2)(x-5)
\(\Leftrightarrow\)15x-51=3x^2-21x+30\(\Leftrightarrow\)3x^2-36x+81=0
\(\Leftrightarrow\)\(\begin{matrix}&\end{matrix}\)\(\left[{}\begin{matrix}9\\3\end{matrix}\right.\) tmđk
16\(\dfrac{x^2-3x+6}{x^2-9}\)=\(\dfrac{1}{x-3}\)(x khác +_3)
\(\Leftrightarrow\)x^2-3x+6=x+3
\(\Leftrightarrow\)x^2-4x+3=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}3loại\\1\end{matrix}\right.\)
vậy x=1 là nghiệm của pt
17 \(\dfrac{3}{x^2-4}\) = \(\dfrac{1}{x-2}+\dfrac{1}{x+2}\)
<=> x + 2 + x - 2 = 3
<=> 2x = 3
<=> x = \(\dfrac{3}{2}\)
TQ:\(S_n=\dfrac{1}{\left(n+n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}\)
Mà theo AM-GM:\(n+\left(n+1\right)\ge2\sqrt{n\left(n+1\right)}\)
\(\Rightarrow S_n\le\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Áp dụng:\(S< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{48}}-\dfrac{1}{\sqrt{49}}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{7}\right)=\dfrac{6}{14}=\dfrac{3}{7}\)
a)\(\sqrt{\dfrac{4x+3}{x+1}}=3< =>\dfrac{4x+3}{x+1}=3^2=9\)
\(=>4x+3=9\left(x+1\right)=9x+9\)
\(=>4x-9x=9-3< =>-5x=6\)
\(=>x=\dfrac{-6}{5}\)
b)\(\dfrac{1}{2}\sqrt{4x-8}-\dfrac{2}{3}\sqrt{x-2}+\sqrt{\dfrac{x-2}{36}}=7\)
\(< =>\dfrac{1}{2}\sqrt{2^2\left(x-2\right)}-\dfrac{2}{3}\sqrt{x-2}+\sqrt{\dfrac{x-2}{6^2}}=7\)
\(< =>\dfrac{1}{2}.2\sqrt{x-2}-\dfrac{2}{3}\sqrt{x-2}+\dfrac{1}{6}\sqrt{x-2}=7\)
\(< =>\left(1-\dfrac{2}{3}+\dfrac{1}{6}\right)\sqrt{x-2}=7\)
\(< =>\dfrac{1}{2}\sqrt{x-2}=7< =>\sqrt{x-2}=7:\dfrac{1}{2}=14\)
\(< =>x-2=14^2=196< =>x=196+2=198\)
a) \(\sqrt{\dfrac{1}{9}.0,4.64}=\sqrt{\dfrac{1}{9}}.\sqrt{0,4}.\sqrt{64}=\dfrac{1}{3}.\dfrac{\sqrt{10}}{5}.8=\dfrac{8\sqrt{10}}{15}\)
b) \(\sqrt{11\dfrac{1}{9}}=\sqrt{\dfrac{100}{9}}=\dfrac{\sqrt{100}}{\sqrt{9}}=\dfrac{10}{3}\)
c) \(\sqrt{\dfrac{1}{44}.2\dfrac{2}{49}}=\sqrt{\dfrac{1}{44}}.\sqrt{\dfrac{100}{49}}=\dfrac{\sqrt{11}}{22}.\dfrac{10}{7}=\dfrac{5\sqrt{11}}{77}\)
d) \(\sqrt{1\dfrac{9}{16}.2\dfrac{1}{4}.2\dfrac{7}{9}}\sqrt{\dfrac{25}{16}.\dfrac{9}{4}.\dfrac{25}{9}}=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{9}{4}}.\sqrt{\dfrac{25}{9}}=\dfrac{5}{4}.\dfrac{3}{2}.\dfrac{5}{3}=\dfrac{25}{8}\)
em mới học lớp 7 thôi ạ