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\(\frac{19n+7}{7n+11}=2\)
\(\Rightarrow x=3\)
Còn cách giải thì k xong mình nói
\(\left(\frac{-.-}{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{ }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}\right)\)
Có: \(A=\frac{19n+7}{7n+11}\)
\(\Leftrightarrow7A=\frac{7\cdot\left(19n+7\right)}{7n+11}=\frac{7\left(19n+209-202\right)}{7n+11}=\frac{19\left(7n+11\right)-202\cdot7}{7n+11}=19-\frac{1414}{7n+11}\)
Mà \(A\in N\Leftrightarrow7A\in N\Leftrightarrow7n+11\inƯ\left(1414\right)=\left\{2;101;7\right\}\)
#)Giải :
Đặt \(A=\frac{19n+7}{7n+11}\)
Để n là số tự nhiên => 19n + 7 chia hết cho 7n + 11
\(\Leftrightarrow7\left(19n+7\right)-19\left(7n+11\right)⋮7n+11\)
\(\Leftrightarrow133n+49-133n-209⋮7n+11\)
\(\Leftrightarrow-160⋮7n+11\)
\(\Leftrightarrow7n+11\in\left\{1;2;4;5;8;10;16;20;32;40;80;160\right\}\)\(\left\{-1;-2;-4;-5;-8;-10;-16;-20;-32;-40;-80;-160\right\}\)
Mà n là số tự nhiên
\(\Rightarrow7n+11\ge11\)
Vậy còn lại các giá trị 16 ; 20 ; 32 ; 40 ; 80 ; 160
Vì các số trên phải chia hết cho 2 => loại các giá trị, còn lại 32
\(\Rightarrow7n+11=32\)
\(\Rightarrow n=3\)
Vậy, khi n = 3 thì A = 2 ( thỏa mãn )
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
bạn ấn vào đúng 0 sẽ ra kết quả, mình giải được rồi dễ lắm
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
Ta có:\(\frac{2748}{1}+\frac{2747}{2}+\frac{2746}{3}+...+\frac{1}{2748}\)
\(=\frac{2747}{2}+1+\frac{2746}{3}+1+...+\frac{1}{2748}+1+1\)
\(=\frac{2749}{2}+\frac{2749}{3}+...+\frac{2749}{2748}+\frac{2749}{2749}\)
Tổng dãy trên là 1 phân số
linh ơi tick mik đ....i.....i.....i....i.....
N=3;-13;-3