\(\left(\frac{1}{4}\right)^3+\frac{-3}{4}.|x-5|=\frac{-9}{2}\)

\(\Rightarrow\frac{1}{64}+\frac{-3}{4}.|x-5|=\frac{-9}{2}\)

\(\Rightarrow\frac{-3}{4}.|x-5|=\frac{-9}{2}-\frac{1}{64}\)

\(\Rightarrow\frac{-3}{4}.|x-5|=\frac{-289}{64}\)

\(\Rightarrow|x-5|=\frac{-289}{64}:\frac{-3}{4}\)

\(\Rightarrow|x-5|=\frac{289}{48}\)

\(\Rightarrow\orbr{\begin{cases}x-5=\frac{289}{48}\\x-5=\frac{-289}{48}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{529}{48}\\x=\frac{-49}{48}\end{cases}}\)

Vậy \(x=\frac{529}{48}\) hoặc \(x=\frac{-49}{48}\)

Ta có : \(\frac{-15}{5}\le x< \frac{22}{11}\)

    \(\Rightarrow-3\le x< 2\)

Vì \(x\in Z\Rightarrow x\in\left\{-3;-2;-1;0;1\right\}\)

a.Ta thấy:\(\frac{3}{4}=\frac{3.9}{4.9}=\frac{27}{36}\)

\(\Rightarrow\frac{3}{4}=\frac{27}{36}\)

b.ta thấy:\(\frac{-4}{5}=\frac{-4.9}{5.9}=\frac{-36}{45};\frac{8}{-9}=\frac{8.\left(-5\right)}{-9.\left(-5\right)}=\frac{-40}{45}\)

Vì\(\frac{-36}{45}>\frac{-40}{45}\)nên\(\frac{-4}{5}\)ko bằng\(\frac{8}{-9}\)

c.Ta thấy:\(\frac{10}{14}=\frac{5}{7};\frac{-15}{-21}=\frac{5}{7}\)

\(\Rightarrow\frac{10}{14}=\frac{-15}{-21}\)

chúc bn hok tốt!!!

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nếu được thì kb luôn nha!!!

\(\left(x+1\right)^2-9=-5\)

\(\Rightarrow\left(x+1\right)^2=-5+9\)

\(\Rightarrow\left(x+1\right)^2=4\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=2^2\\\left(x+1\right)^2=\left(-2\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

Vậy x = 1 hoặc x = -3

\(\left(x+1\right)^2-9=-5\)

\(\Leftrightarrow\left(x+1\right)^2=-5+9\)

\(\Leftrightarrow\left(x+1\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=2^2\\\left(x+1\right)^2=\left(-2\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2-1\\x=-2-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

Vậy x=1 hoặc x=-3

\(a)\) Ta có : 

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(............\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\)\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow\)\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\)\(A< 1+1-\frac{1}{100}\)

\(\Rightarrow\)\(A< 2-\frac{1}{100}< 2\)

\(\Rightarrow\)\(A< 2\) ( đpcm ) 

Vậy \(A< 2\)

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