- Tim cac so nguyen x,y biet \(\frac{1}{18}< \frac{x}{12}< \frac{y}{9}< \frac{1}{4}\)
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1/18 < x/12 < y/9 < 1/4.
Ta quy dong mau len co mau chung la 36: 2/36 < x.3/36 < y.4/36 < 9/36.
Suy ra: Vi 2<x<y<9 nen phai bang 3;4;5;6;7;8:
x.3 3 4 5 6 7 8
x 1 loai loai 2 loai loai
y.4 3 4 5 6 7 8
y loai 1 loai loai loai 2
Suy ra ta co 3 truong hop:
TH1: x=1;y=1: 2/36 < 1.3/36 < 1.4/36 < 9/36
TH2: x=2;y=2: 2/36 < 2.3/36 < 2.4/36 < 9/36
TH3: x=1;y=1: 2/36 < 1.3/36 < 2.4/36 < 9/36
4,58 - (3,125 + 1,105) < x < 9,1 - ( 6,85 - 2,75)
=> 4,58 - 4,23 < x < 9,1 - 4,1
=> 0,35 < x < 5
=> x \(\in\){1; 2; 3; 4}
4.58-(3.125+1.105)<x<9.1-(6.85-2.75)
4.58-4.28<x<9.1-4.1
0.35<x<5
x\(\in\)1.2.3.4
\(\frac{1}{15}<\frac{x}{12}<\frac{x}{9}<\frac{1}{4}\)
\(\Rightarrow\frac{2}{36}<\frac{3x}{36}<\frac{4y}{36}<\frac{9}{36}\)
Ta có:\(\frac{2}{36}<\frac{3x}{36}<\frac{9}{36}\)
\(\Rightarrow\)\(2<3x<9\)
\(\Rightarrow\)\(\frac{2}{3}\)<x<3
\(\Rightarrow1\le\)x\(<3\)
\(\Rightarrow x\in\left\{1,2,3\right\}\)
\(x=1\Rightarrow\frac{3}{36}<\frac{4y}{36}<\frac{9}{36}\)\(\Rightarrow\)\(3<4y<9\)
\(\Rightarrow\frac{3}{4}\)\(<\)x\(<\frac{9}{4}\)
\(\Rightarrow\)\(1\)\(\le\)x\(\le2\)
\(x=2\) và \(x=3\) tương tự
\(a)\) \(\frac{-11}{12}< \frac{x}{12}< \frac{-3}{4}\)
\(\Leftrightarrow\)\(\frac{-11}{12}< \frac{x}{12}< \frac{-9}{12}\)
\(\Leftrightarrow\)\(-11< x< -9\)
\(\Rightarrow\)\(x=-10\)
Quy đồng: mẫu số chung : 72
\(\frac{1}{18}=\frac{4}{72}\)
\(\frac{x}{12}=\frac{x}{72}\)
\(\frac{y}{9}=\frac{y}{72}\)
\(\frac{1}{4}=\frac{18}{72}\)
=>\(\frac{1}{12}=\frac{6}{72}\)
=>\(\frac{1}{9}=\frac{8}{72}\)
so sánh: \(\frac{1}{12}< \frac{1}{9}\) vì \(\frac{6}{72}< \frac{8}{72}\)
\(\Rightarrow x=1\) ; \(y=1\)