Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7:\left(\frac{3}{5}\right)^2\)
\(x=\frac{3^7}{7^7}:\frac{3^2}{5^2}\)
\(x=\frac{3^7.5^2}{7^7.3^2}\)
\(x=\frac{3^5.5^2}{7^7}=\frac{6075}{823543}\)
\(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{2}{5}< x-\frac{7}{5}< \frac{3}{5}\\\frac{2}{5}>x-\frac{7}{5}>\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{2}{5}+\frac{7}{5}< x-\frac{7}{5}+\frac{7}{5}< \frac{3}{5}+\frac{7}{5}\\\frac{2}{5}+\frac{7}{5}>x-\frac{7}{5}+\frac{7}{5}>\frac{3}{5}+\frac{7}{5}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{9}{5}< x< 2\\\frac{9}{5}>x>2\end{cases}}\)
Vậy \(\frac{9}{5}< x< 2\) hoặc \(\frac{9}{5}>x>2\)
Chúc bạn học tốt ~
1) \(\frac{x-1}{5}=\frac{x+2}{7}\)
\(\Rightarrow7x-7=5x+10\)
=> 7x - 5x = 10 + 7
2x = 17
x = 8,5
b) \(\frac{x-1}{7}=\frac{5}{x+1}\)
=> x2 - 1 = 35
x2 = 34
\(\Rightarrow x=\sqrt{34};x=-\sqrt{34}\)
c) \(\frac{x+1}{x-2}=\frac{x+2}{x+3}\)
=> x2 + 4x + 3 = x2 - 4
=> x2 - x2 + 4x + 3 = -4
4x + 3 = - 4
4x = -7
x = -7/4
a ) \(1\frac{1}{2}+x=\frac{3}{7}-7\)
\(\frac{3}{2}+x=-\frac{46}{7}\)
\(x=-\frac{46}{7}-\frac{3}{2}\)
\(x=-\frac{113}{14}\)
1) Chỉ tìm được Max thôi nhé
a) \(C=\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\le\frac{4}{5}+\frac{20}{8}=\frac{33}{10}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|3x+5\right|=0\\\left|4y+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{5}{3}\\y=-\frac{5}{4}\end{cases}}\)
b) \(E=\frac{2}{3}+\frac{21}{\left(x+3y\right)^2+5\left|x+5\right|+14}\le\frac{2}{3}+\frac{21}{14}=\frac{13}{6}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+3y\right)^2=0\\5\left|x+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{5}{3}\end{cases}}\)
2) Thì chỉ tìm được GTNN thôi nhé
a) \(A=5+\frac{-8}{4\left|5x+7\right|+24}\ge5-\frac{8}{24}=\frac{14}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(4\left|5x+7\right|=0\Rightarrow x=-\frac{7}{5}\)
Vậy Min(A) = 14/3 khi x = -7/5
b) \(B=\frac{6}{5}-\frac{14}{5\left|6y-8\right|+35}\ge\frac{6}{5}-\frac{14}{35}=\frac{4}{5}\left(\forall y\right)\)
Dấu "=" xảy ra khi: \(5\left|6y-8\right|=0\Rightarrow x=\frac{4}{3}\)
Vậy Min(B) = 4/5 khi x = 4/3