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\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
\(C\in\left\{1;3\right\}\)
\(D\in\left\{1;4\right\}\)
\(E\in\left\{2;3\right\}\)
\(F\in\left\{2;4\right\}\)
\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)
1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\)
\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\)
\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)
a) \(\frac{x-2}{3}=\frac{x+1}{4}\)
=> (x - 2).4 = 3.(x + 1)
=> 4x - 8 = 3x + 3
=> 4x - 3x = 3 + 8
=> x = 11
Vậy x = 11
b) \(2.\left(x+3\right)-\frac{1}{2}=x-1\)
=> \(2x+6-\frac{1}{2}=x-1\)
=> \(2x+\frac{11}{2}=x-1\)
=> \(2x-x=-1-\frac{11}{2}\)
=> \(x=-\frac{13}{2}\)
Vậy \(x=-\frac{13}{2}\)
a) ta có: \(\frac{1}{x}-\frac{y}{6}=\frac{1}{3}\)<=> \(\frac{1}{x}=\frac{1}{3}+\frac{y}{6}\)
<=> \(\frac{1}{x}=\frac{2+y}{6}\)<=> \(x\left(2+y\right)=6\)
Mà x, y nguyên => x và y+2 \(\inƯ_{\left(6\right)}=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
thay vào ta tìm được các cặp x,y.
b) Ta có: \(\frac{x}{2}+\frac{3}{y}=\frac{5}{4}\)<=> \(\frac{3}{y}=\frac{5}{4}-\frac{x}{2}\)
<=> \(\frac{3}{y}=\frac{5-2x}{4}\)
<=> \(y\left(5-2x\right)=12\)
vì x,y nguyên , 5-2x luôn lẻ => 5-2x \(\inƯ_{\left(12\right)}=\left\{\pm1;\pm3\right\}\)
Thay vào ta tìm được các cặp x,y.
\(\frac{3x-11}{2}-\frac{x-3}{3}=\frac{1}{6}\)
\(\frac{3\times\left(3x-11\right)}{3\times2}-\frac{2\times\left(x-3\right)}{2\times3}=\frac{1}{6}\)
\(\frac{9x-33}{6}-\frac{2x-6}{6}=\frac{1}{6}\)
\(\frac{\left(9x-33\right)-\left(2x-6\right)}{6}=\frac{1}{6}\)
\(9x-33-2x+6=1\)
\(\left(9x-2x\right)-\left(33-6\right)=1\)
\(7x-27=1\)
\(7x=1+27\)
\(7x=28\)
\(x=\frac{28}{7}\)
\(x=4\)
Chúc bạn học tốt
\(PT\Leftrightarrow\frac{3.\left(3x-11\right)-2.\left(x-3\right)}{6}=\frac{1}{6}\)
<=> 3.(3x - 11) - 2.(x - 3) = 1
<=> 9x - 33 - 2x + 6 = 1
<=> 7x = 28
<=> x = 4
55/21; 134/43; 116/37; 55/21
theo phân số: 55/21;134/43;116/37;77/21
theo số thập phân:2.(619047); 3.(11627...); 3.(135); 4.(0526...)
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