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Câu a hình như sai đề mk sửa nha
a)\(A=\left(2x+\frac{1}{3}\right)^4-1\)
Vì \(\left(2x+\frac{1}{3}\right)^4\ge0\)
Suy ra:\(\left(2x+\frac{1}{3}\right)^4-1\ge-1\)
Dấu = xảy ra khi \(2x+\frac{1}{3}=0\)
\(2x=-\frac{1}{3}\)
\(x=-\frac{1}{6}\)
Vậy Min A=-1 khi \(x=-\frac{1}{6}\)
b)\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)
\(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)
Vì \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)
Suy ra:\(3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le3\)
Dấu = xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\)
\(\frac{4}{9}x=\frac{2}{15}\)
\(x=\frac{3}{10}\)
Vậy Max B=3 khi \(x=\frac{3}{10}\)
Bài 1:
\(a,A=\frac{-25}{28}.0,21=\frac{-25}{28}.\frac{21}{100}=\frac{-25.21}{28.100}=\frac{-1.25.3.7}{4.7.25.4}=\frac{-1.3}{4.4}=\frac{-3}{16}\)
\(b,B=\left(\frac{13}{24}-\frac{29}{30}\right):\left(-10,2\right)=\left(\frac{65}{120}-\frac{116}{120}\right):\frac{-51}{5}=\frac{-51}{120}.\frac{5}{-51}=\frac{-51.5}{120.\left(-51\right)}=\frac{-51.5}{5.24.\left(-51\right)}=\frac{1}{24}\)
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49.\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
Đặt \(A=1-3+3^2-3^3+...+\left(-3^x\right)\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...+\left(-3^{x+1}\right)\)
\(\Rightarrow3A+A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow4A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow A=\frac{1+\left(-3^{x+1}\right)}{4}\)
\(\Rightarrow\frac{1+\left(-3^{x+1}\right)}{4}=\frac{1-9^{1006}}{4}\)
\(\Rightarrow-3^{x+1}=-9^{1006}\)
\(\Rightarrow-3^{x+1}=-3^{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2012-1\)
\(\Rightarrow x=2011\)
Vậy : \(x=2011\)
Bài 3:
a) Ta có: \(1.25\cdot\left(-3\frac{3}{8}\right)\)
\(=\frac{5}{4}\cdot\frac{-27}{8}\)
\(=\frac{-135}{32}\)
b) Ta có: \(\frac{-9}{34}\cdot\frac{17}{4}\)
\(=\frac{-9}{4}\cdot\frac{17}{34}\)
\(=-\frac{9}{4}\cdot\frac{1}{2}\)
\(=-\frac{9}{8}\)
c) Ta có: \(-\frac{20}{41}\cdot\frac{-4}{5}\)
\(=\frac{20}{41}\cdot\frac{4}{5}\)
\(=\frac{16}{41}\)
d) Ta có: \(\frac{-6}{7}\cdot\frac{21}{2}\)
\(=-\frac{6}{2}\cdot\frac{21}{7}\)
\(=-3\cdot3=-9\)
Bài 4:
a) Ta có: \(-\frac{5}{2}\cdot\frac{3}{4}\)
\(=-\frac{5\cdot3}{2\cdot4}=\frac{-15}{8}\)
b) Ta có: \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=-\frac{21}{5}:\frac{14}{5}\)
\(=-\frac{21}{5}\cdot\frac{5}{14}\)
\(=-\frac{21}{14}=-\frac{3}{2}\)
c) Ta có: \(1.8:\left(-\frac{3}{4}\right)\)
\(=\frac{9}{5}:\frac{-3}{4}\)
\(=\frac{9}{5}\cdot\frac{4}{-3}\)
\(=-\frac{12}{5}\)
d) Ta có: \(\frac{17}{15}:\frac{4}{3}\)
\(=\frac{17}{15}\cdot\frac{3}{4}\)
\(=\frac{17}{20}\)
Cái này mk làm rồi
Ừm