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B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
Bài 2:
\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-2}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3.\left(\dfrac{2}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^5\)
Vì \(\dfrac{2}{5}\ne\pm1;\dfrac{2}{5}\ne0\) nên \(x>5\)
Vậy \(x>5\) thoả mãn yêu cầu đề bài.
Chúc bạn học tốt!!!
Bài 1:
\(C=\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\)
\(C=\left(\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right).....\left(\dfrac{1}{\left(100-1\right)\left(100+1\right)}\right)\)
\(C=\dfrac{1}{1.3}\dfrac{1}{2.4}.....\dfrac{1}{99.101}=\dfrac{1}{101!}\)
Chúc bạn học tốt!!!
1/ \(4\left(\dfrac{-1}{2}\right)^3+\dfrac{1}{2}:5\)
\(=4.\dfrac{-1}{8}+\dfrac{1}{2}.\dfrac{1}{5}\)
\(=\dfrac{-1}{2}+\dfrac{1}{10}\)
\(=\dfrac{-5}{10}+\dfrac{1}{10}\)
\(=\dfrac{-4}{10}\)
\(=\dfrac{-2}{5}\)
2/ \(17\dfrac{1}{5}:\left(-\dfrac{5}{7}\right)-2\dfrac{1}{5}.\left(-\dfrac{7}{5}\right)\)
\(=\dfrac{86}{5}.\left(\dfrac{-7}{5}\right)-\dfrac{11}{5}.\left(\dfrac{-7}{5}\right)\)
\(=\dfrac{-7}{5}.\left(\dfrac{86}{5}-\dfrac{11}{5}\right)\)
\(=\dfrac{-7}{5}.15\)
\(=-21\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a, \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\left(\dfrac{5}{3}-\dfrac{1}{4}\right).\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
b, \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(-\dfrac{1}{2}\right)^3=2:-\dfrac{1}{8}=2.-8=-16\)
\(a.\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\left(\dfrac{5}{3}-\dfrac{1}{4}\right).\left(\dfrac{1}{20}\right)^2\)
\(=\left(\dfrac{17}{12}\right).\dfrac{1}{400}\)
\(=\dfrac{17}{4800}\)
\(b.2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3\)
\(=2:\left(-\dfrac{1}{6}\right)^3\)
\(=2:\left(-\dfrac{1}{216}\right)\)
\(=\left(-432\right)\)
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)
= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)
= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)
= \(\dfrac{215}{1}=215\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)
= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)
= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)
= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)
= \(\dfrac{300}{2}=150\)
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
B=\(\dfrac{1}{2}:\left(-1\dfrac{1}{2}\right):1\dfrac{1}{3}:....:\left(-1\dfrac{1}{100}\right)\)
=\(\dfrac{1}{2}:\dfrac{-3}{2}:\dfrac{4}{3}:....:\dfrac{-101}{100}\)
=\(\dfrac{1}{2}.\dfrac{-2}{3}.\dfrac{3}{4}........\dfrac{-100}{101}\)
=\(\dfrac{1.\left(-2\right).3......\left(-100\right)}{2.3.4...........101}\)
=\(\dfrac{1}{101}\)