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Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
làm tiếp cái trước(ấn nhầm)
\(x=\frac{5}{42}-\frac{15}{28}\)
\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)
\(x=\frac{20}{168}-\frac{90}{168}\)
\(x=\frac{-70}{168}\)
\(x=\frac{-5}{12}\)
2.
1.
\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)
\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
\(\left(-\frac{5}{9}\right).\frac{3}{11}+\left(-\frac{13}{28}\right).\frac{3}{11}\)
\(=\frac{3}{11}.\left(-\frac{5}{9}+\left(-\frac{13}{28}\right)\right)\)
\(=\frac{3}{11}.\frac{-257}{252}\)
\(=-\frac{257}{924}\)
Ta có : \(C=\left(1+\frac{11}{13}\right).\left(1+\frac{11}{28}\right).\left(1+\frac{11}{45}\right)......\left(1+\frac{11}{220}\right)\)
\(=\frac{24}{13}.\frac{39}{28}.\frac{56}{45}......\frac{231}{220}\)
\(=\frac{2.12}{1.13}.\frac{3.13}{2.14}.\frac{4.14}{3.15}......\frac{11.21}{10.22}\)
\(=\frac{2.12.3.13.4.14.....11.21}{1.13.2.14.3.15.....10.22}\)
\(=\frac{\left(2.3.4.....11\right).\left(12.13.14.....21\right)}{\left(1.2.3.....10\right).\left(13.14.15.....22\right)}\)
\(=\frac{11.12}{1.22}=\frac{12}{1.2}=6\)
Vậy C = 6
Nguyễn Thị Tiểu Ngân Lớp 7a thcs Dĩnh Kế phải ko ?