\(0< a_i< 10,\forall i=1,2,3,...,n\)

\(\overline{a_1a_2}+\overline{a_2a_3}+\overline{a_3a_4}+...+\overline{a_{n-1}a_n}+\overline{a_na_1}\)

\(=10a_1+a_2+10a_2+a_3+10a_3+a_4+...+10a_{n-1}+a_n+10a_n+a_1\)

\(=11a_1+11a_2+11a_3+...+11a_{n-1}+11a_n\)

\(=11\left(a_1+a_2+...+a_{n-1}+a_n\right)\left(đpcm\right)\)

em đăng lại câu hỏi vào box Toán học nhé.

\(a,A=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{3}.\sqrt{2}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(=3-2=1\)

\(b,B=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{\dfrac{9}{4}.8}-2\sqrt{\dfrac{9}{4}.12}+\sqrt{\dfrac{9}{4}.20}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3.\dfrac{3}{4}\sqrt{8}-2.\dfrac{3}{4}\sqrt{12}+\dfrac{3}{4}\sqrt{20}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{\dfrac{3}{4}\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right)}\)

\(=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(A=\dfrac{x}{x-5}-\dfrac{10x}{x^2-25}-\dfrac{5}{x+5}\left(ĐKXĐ:x\ne\pm5\right)\)

\(=\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{10x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+5x-10x-5x+25}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)

`a)`\(A=\dfrac{x}{x-1}+\dfrac{1}{x+2}-\dfrac{3x}{x^2+x-2}\)

\(A=\dfrac{x}{x-1}+\dfrac{1}{x+2}-\dfrac{3x}{\left(x-1\right)\left(x+2\right)}\)

\(A=\dfrac{x\left(x+2\right)+\left(x-1\right)-3x}{\left(x-1\right)\left(x+2\right)}\)

\(A=\dfrac{x^2+2x+x-1-3x}{\left(x-1\right)\left(x+2\right)}\)

\(A=\dfrac{x^2-1}{\left(x-1\right)\left(x+2\right)}\)

\(A=\dfrac{x+1}{x+2}\)

`b)`\(S=A.B\)

\(S=\dfrac{x+1}{x+2}.\dfrac{x+3}{x+1}\)

\(S=\dfrac{x+3}{x+2}\)

\(S=\dfrac{x+2+1}{x+2}=1+\dfrac{1}{x+2}\)

Ta có:\(x\ge0\Rightarrow x+2\ge2\)

\(\Rightarrow S\le1+\dfrac{1}{2}=\dfrac{3}{2}\)

Vậy \(Max_S=\dfrac{3}{2}\) khi \(x=0\)

332

`a)`\(M=A.B\)

\(M=\left(\dfrac{x+2}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)

\(M=\left(\dfrac{-\left(x+2\right)^2-4x^2-\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-3}{2x-x^2}\right)\)

\(M=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)

\(M=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)

\(M=\dfrac{-4x\left(2x-x^2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(M=\dfrac{4x^2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(M=\dfrac{4x^2}{x-3}\)

`b)`\(\left|x-7\right|=4\)

 \(\Leftrightarrow\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)

`@` Với `x=11`\(\Rightarrow M=\dfrac{121}{2}\)

`@` Với `x=3` `=>` `M` không xác định

`c)`\(A>0\)

\(\Leftrightarrow\dfrac{4x^2}{x-3}>0\)

`<=>x-3>0`

`<=>x>3`

.

`a)`\(S=1+A:B\)

\(S=1+\left(\dfrac{x^3-2x^2}{x^3-x^2+x}\right):\left(\dfrac{x+1}{x^3+1}+\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right)\)

\(S=1+\left(\dfrac{x^2-2x}{x^2-x+1}\right):\left(\dfrac{x+1+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)

\(S=1+\dfrac{x^2-2x}{x^2-x+1}:\dfrac{4x-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(S=1+\dfrac{x^2-2x}{x^2-x+1}.-\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{2\left(x^2-2x\right)}\)

\(S=1-\dfrac{x+1}{2}\)

`b)`\(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=-\dfrac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)

`@`Với `x=2` \(\Rightarrow S=1+\dfrac{2+1}{2}=1+\dfrac{3}{2}=\dfrac{5}{2}\)

`@`Với `x=-1/2` \(\Rightarrow S=1+\dfrac{-\dfrac{1}{2}+1}{2}=\dfrac{5}{4}\)

`c)`\(S=1-\dfrac{x+1}{2}\)

Để `S` nguyên thì \(x+1⋮2\) hay `x` thuộc các số lẻ

giúp mình với ạ

`a)`\(P=A:B\)

\(P=\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\)

\(P=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(P=\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-x^2}{\left(x-1\right)\left(x+1\right)}\)

\(P=-\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)

`b)`\(P=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2}\)

\(\Leftrightarrow2\left(2x^2+2\right)=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow4x^2+4=x^2-1\)

\(\Leftrightarrow3x^2=-5\) ( vô lý )

Vậy không có giá trị `x` thỏa mãn `P=1/2`

help^^

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