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\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}\le1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
<=> \(\frac{41}{36}\le\frac{x}{36}\le\frac{35}{24}\)
<=> \(\frac{82}{72}\le\frac{2x}{72}\le\frac{105}{72}\)
<=> \(82\le2x\le105\)
<=> \(41\le x\le52,5\)
Do \(x\in N\)nên \(x=\left\{x\in N|41\le x\le52,5\right\}\)
\(\dfrac{1}{4}+\dfrac{8}{9}\le\dfrac{x}{36}\le1-\left(\dfrac{3}{8}-\dfrac{5}{6}\right)\\ \Rightarrow\dfrac{41}{36}\le\dfrac{x}{36}\le\dfrac{35}{24}\\ \Rightarrow\dfrac{82}{72}\le\dfrac{2x}{72}\le\dfrac{105}{72}\\ \Rightarrow41\le x< 51,5\)
\(\left(\frac{1}{4}.x-1\right)+\left(\frac{5}{6}.x-2\right)-\left(\frac{3}{8}.x+1\right)=\frac{9}{2}\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1=\frac{9}{2}\)
\(\Rightarrow\frac{1}{4}x+\frac{5}{6}x-\frac{3}{8}x=\frac{9}{2}+1+2+1\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\div\frac{17}{24}=12\)
Ta có quy luật:
1+1=12 = 1
2+2=22 = 4
3+3 = 32 = 9
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9 + 9 = 92 = 81
\(\dfrac{1}{4}+\dfrac{8}{9}\le\dfrac{x}{36}< 1-\left(\dfrac{3}{8}-\dfrac{5}{6}\right)\)
\(\Rightarrow\dfrac{41}{36}\le\dfrac{x}{36}< 1-\left(-\dfrac{11}{24}\right)\)
\(\Rightarrow\dfrac{41}{36}\le\dfrac{x}{36}< 1+\dfrac{11}{24}\)
\(\Rightarrow\dfrac{41}{36}\le\dfrac{x}{36}< \dfrac{35}{24}\)
\(\Rightarrow\dfrac{82}{72}\le\dfrac{2x}{72}< \dfrac{105}{72}\)
\(\Rightarrow82\le2x< 105\)
\(\Rightarrow41\le x< \dfrac{105}{2}\)
\(\dfrac{1}{4}+\dfrac{8}{9}\le\dfrac{x}{36}< 1-\left(\dfrac{3}{8}-\dfrac{5}{6}\right)\)
\(\dfrac{\Leftrightarrow41}{36}\le\dfrac{x}{36}< \dfrac{35}{24}\)
\(\dfrac{\Leftrightarrow82}{72}\le\dfrac{2x}{72}< \dfrac{105}{72}\)
\(\Rightarrow82\le2x< 105\)
\(\Rightarrow x=\left\{41;42;43;44;45;46;47;48;49;50;51;52\right\}\)