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B1:
\(A=\left(x+2020\right)^4+\left|y-2019\right|-2018\)
+Có: \(\left(x+2020\right)^4\ge0với\forall x\\\left|y-2019\right|\ge0với\forall y\\\Rightarrow \left(x+2020\right)^4+\left|y-2019\right|-2018\ge-2018\\ \Leftrightarrow A\ge-2018 \)
+Dấu "=" xảy ra khi
\(\left(x+2020\right)^4=0\\ \Leftrightarrow x=-2020\)
\(\left|y-2019\right|=0\\ \Leftrightarrow y=2019\)
+Vậy \(A_{min}=-2018\) khi \(x=-2020,y=2019\)
\(A=\left(-2\right)\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{214}\right)\)
\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{215}{214}=215\)
\(B=\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)....\left(-1\frac{1}{299}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{300}{299}=\frac{300}{2}=150\)
\(C=-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{333333}{424242}\right)\)
\(=-\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=-\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}.\frac{4}{21}=-11\)
a) \(\left(0,25+\frac{3}{4}:1,25+1\frac{1}{3}:2\right)\)\(=\left(\frac{1}{4}+\frac{3}{4}\right):\frac{5}{4}+\frac{4}{3}:2\)\(=\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\)
b) \(2^3+3\left(\frac{-3}{2}\right)^0.\left(\frac{1}{2}\right)^2.4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)\(=8+3.1.\frac{1}{4}.4+\left[4:\frac{1}{2}\right]:8\)
\(=8+3+8:8=\frac{19}{8}\)
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
Thật ra tui cũng không rõ lắm đâu. Cậu thử nhân A với \(\dfrac{2019}{2020}\)rồi lại cộng lại với A thử coi nào <Chú Ý : chưa chắc đã đúng >
x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)
\(=\frac{7}{12}\)
b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
iem chỉ biết làm câu đầu , NHƯNG KO BÍT có ĐUG HAY KO
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2018\cdot2019}{2\cdot3\cdot4\cdot..\cdot2019\cdot2020}\)
\(A=\frac{1}{2020}\)
Với \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\) , ta có : \(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}=\frac{1}{2020}\)
Ta có \(7A=\frac{7}{2020}\) , \(9A=\frac{9}{2020}\) , \(1+A=\frac{2021}{2020}\)
\(\frac{1+7A}{1+9A}=\frac{1+\frac{7}{2020}}{1+\frac{9}{2020}}=\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)
Ta thấy \(\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)có tử kém mẫu \(\frac{2}{2020}\)đơn vị và không thể rút gọn được nữa .
\(\Rightarrow\frac{1+7A}{1+9A}\)là p/s tối giản.