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a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)

\(3\frac{1}{5}x+125x=178\)
=> 3,2x + 125x = 178
=> 128,2x = 178
\(\Rightarrow x=\frac{890}{641}\)
\(3\frac{1}{5}x+125x=178 \)
\(\frac{16}{5}x+125x=178\)
\(x.\left(\frac{16}{5}+125\right)=178\)
\(x.\frac{641}{5}=178\)
\(x=178:\frac{641}{5}\)
\(x=\frac{890}{641}\)
vậy \(x=\frac{890}{641}\)

Ta có \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
Ta thấy \(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}>0\)suy ra \(3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Khi đó \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 3\)

Ta có
A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)
Đặt C = 1 + 7 + 72 + 73+...+711
7C = 7 + 72 + 73 + ... + 711 + 712
=> 6C = 712 - 1
C = \(\dfrac{7^{12}-1}{6}\)
Đặt D = 1 + 7 + 72 + 73+...+710
7D = 7 + 72 + 73 + ... + 710 + 711
=> 6D = \(7^{11}-1\)
D = \(\dfrac{7^{11}-1}{6}\)
=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)
A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)
A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)
A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003
Lại có:
B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\
Đặt H = \(1+3+3^2+3^3+...+3^{11}\)
3H = \(3+3^2+3^3+...+3^{12}\)
=> 2H = \(3^{12}-1\)
H = \(\dfrac{3^{12}-1}{2}\)
Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)
3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)
=> 2Q = \(3^{11}-1\)
Q = \(\dfrac{3^{11}-1}{2}\)
=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)
B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)
B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)
B = \(\dfrac{3^{12}-1}{3^{11}-1}\)
B = 3, 00001129
Vì 7, 000000003 > 3, 00001129
=> A > B
Vậy A > B

\(A=1+7+7^2+7^3+...+7^{2016}\)
\(\Rightarrow7A=7\left(1+7+7^2+7^3+...+7^{2016}\right)\)
\(7A=7+7^2+7^3+7^4+...+7^{2017}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2017}\right)-\left(1+7+7^2+...+7^{2016}\right)\)
\(\Rightarrow6A=7^{2017}-1\)
\(\Rightarrow A=\dfrac{7^{2017}-1}{6}\)

Kết quả là 2870.
Giải nhanh bằng công thức tổng bình phương:
\(1^{2} + 2^{2} + \hdots + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)
Với \(n = 20\):
\(\frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870.\)
Ta có biểu thức:
\(1\times1+2\times2+3\times3+\ldots+20\times20=1^2+2^2+3^2+\ldots+20^2\)
Đây là tổng các số chính phương từ 1 đến 20.
Áp dụng công thức tổng bình phương:
\(1^2+2^2+3^2+\ldots+n^2=\frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)
Thay \(n = 20\):
\(\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870\)

\(\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+1+\frac{7}{17}}\div\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+\frac{7}{7}+\frac{7}{17}}\div\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{3\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{7\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\div\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{3}{7}\div\frac{2}{7}\)
\(=\frac{3}{7}\times\frac{7}{2}=\frac{3}{2}\)

1) 4824 - 4824 : 24 - 12 = 4824 - 201 - 12 = 4623 - 12 = 4611

1)\(\left(2\dfrac{3}{17}-2\dfrac{3}{5}\right)+\left(-2\dfrac{3}{17}-1\dfrac{2}{5}\right)\)
=\(\dfrac{37}{17}-\dfrac{13}{5}+\left(-\dfrac{37}{17}\right)-\dfrac{7}{5}\)
=\(\left[\dfrac{37}{17}+\left(-\dfrac{37}{17}\right)\right]-\left(\dfrac{13}{5}+\dfrac{7}{5}\right)\)
=\(0-4=-4\)
2)\(\left(2\dfrac{7}{15}-3\dfrac{3}{7}\right)-\left(-\dfrac{9}{21}+3\dfrac{7}{15}\right)\)
=\(2\dfrac{7}{15}-3\dfrac{3}{7}+\dfrac{9}{21}-3\dfrac{7}{15}\)
=\(\left(2\dfrac{7}{15}-3\dfrac{7}{15}\right)+\left(-3\dfrac{3}{7}+\dfrac{9}{21}\right)\)
=\(-1+\left(-\dfrac{24}{7}+\dfrac{9}{21}\right)\)
=\(\left(-1\right)+\left(-3\right)\)
=-4
3)\(\left(2\dfrac{7}{19}+5\dfrac{3}{7}\right)+\left(-\dfrac{14}{38}+1\dfrac{4}{7}\right)\)
\(=2\dfrac{7}{19}+5\dfrac{3}{7}+\left(-\dfrac{14}{38}\right)+1\dfrac{4}{7}\)
\(=\left(5\dfrac{3}{7}+1\dfrac{4}{7}\right)+\left[2\dfrac{7}{19}+\left(-\dfrac{14}{38}\right)\right]\)
\(=7+\left[\dfrac{45}{19}+\left(-\dfrac{14}{38}\right)\right]\)
\(=7+2=9\)
Hai câu(2),(3)mình làm bằng cách cộng trừ hỗn số cho nhanh nếu bạn không làm cách đó thì đổi ra p/s làm cũng được
\(-178-\left(3x+1\right)=\left(-3\right).\left(-7\right)\)
\(-178-\left(3x+1\right)=21\)
\(3x+1=-178-21\)
\(3x+1=-199\)
\(3x=-199-1\)
\(3x=-200\)
\(x=-200:3\)
\(x=\frac{-200}{3}\)
\(Vậy\) \(x\) \(=\) \(\frac{-200}{3}\)