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\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
a) Để \(\frac{11}{\sqrt{x}-5}\)nhận giá trị nguyên thì \(\sqrt{\text{x}}-5\inƯ\left(11\right)\)(DK : \(0\le x\ne25\))
Vì \(\sqrt{\text{x}}-5\ge-5\)nên ta có :
\(\sqrt{x}-5\in\left\{-1;1;11\right\}\)\(\Rightarrow\sqrt{x}\in\left\{4;6;16\right\}\Rightarrow x\in\left\{16;36;256\right\}\)
b) \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)(DK : \(0\le x\ne9\))
Để B nhận giá trị nguyên thì \(\sqrt{x}-3\inƯ\left(4\right)\)
Vì \(\sqrt{\text{x}}-3\ge-3\)nên ta có :
\(\sqrt{\text{x}}-3\in\left\{-2;-1;1;2;4\right\}\)\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)
\(\frac{5+x}{4-x}=\frac{1}{2}\)
\(\Rightarrow10+2\cdot x=4-x\)
\(\Rightarrow6=-3\cdot x\)
\(\Rightarrow x=-2\)
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{10}\)\(\Leftrightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{10}\)
\(\Leftrightarrow\frac{11}{10}x=\frac{-33}{10}\)\(\Leftrightarrow x=\frac{-33}{10}:\frac{11}{10}=\frac{-33}{10}.\frac{10}{11}=-3\)
Vậy \(x=-3\)
\(\frac{1}{2}\cdot x+\frac{3}{5}\cdot x=-\frac{33}{10}\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)\cdot x=-\frac{33}{10}\)
\(\left(\frac{5}{10}+\frac{6}{10}\right)\cdot x=-\frac{33}{10}\)
\(\frac{11}{10}\cdot x=-\frac{33}{10}\)
\(x=-\frac{33}{10}:\frac{11}{10}=-\frac{33}{10}\cdot\frac{10}{11}\)
\(x=-\frac{33}{11}=-3\)
\(\frac{5}{2}.x-\frac{1}{3}.x+2=\frac{3}{2}\)
\(\frac{5}{2}.x-\frac{1}{3}.x=\frac{3}{2}-2\)
\(\left(\frac{5}{2}-\frac{1}{3}\right).x=-\frac{1}{2}\)
\(\frac{13}{6}.x=-\frac{1}{2}\)
\(x=-\frac{1}{2}:\frac{13}{6}\)
\(x=-\frac{3}{13}\)
\(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+5}{14}+\frac{x+4}{15}\)
\(\Rightarrow\frac{x+1}{18}+1+\frac{x+2}{17}+1=\frac{x+5}{14}+1+\frac{x+4}{15}+1\)
\(\Rightarrow\frac{x+1}{18}+\frac{18}{18}+\frac{x+2}{17}+\frac{17}{17}=\frac{x+5}{14}+\frac{14}{14}+\frac{x+4}{15}+\frac{15}{15}\)
\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}=\frac{x+19}{14}+\frac{x+19}{15}\)
\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{14}-\frac{x+19}{15}=0\)
\(\Rightarrow\left(x+19\right).\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\text{Mà }\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{14}-\frac{1}{15}\right)\ne0\text{ nên: }x+19=0\Rightarrow x=-19\)
Ta có : \(M=\frac{x-5}{\left|5-x\right|}=1\)
\(\Rightarrow x-5=\left|5-x\right|\)
Vì x - 5 là số đối của 5 - x
\(\Rightarrow5-x\le0\Rightarrow x\ge5\)
\(\Rightarrow x-5=5-x\)
\(\Leftrightarrow x+x=5+5\)
\(\Leftrightarrow x=5\)