Lời giải:

$\frac{5}{14}+\frac{-6}{17}+\frac{17}{41}+\frac{9}{14}+\frac{-24}{41}$

$=(\frac{5}{14}+\frac{9}{14})+\frac{-6}{17}+(\frac{17}{41}+\frac{-24}{41})$
$=\frac{14}{14}+\frac{-6}{17}+\frc{-7}{41}$

$=1-(\frac{6}{17}+\frac{7}{41})=1-\frac{365}{697}=\frac{332}{697}$

C

c nhá bạn mệt quá

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

Theo Co-si, ta có: 

\(a^3+b^3+c^3>=3\sqrt[3]{a^3\cdot b^3\cdot c^3}=3bac\)

Dấu '=' xảy ra khi a=b=c

\(a^2=2\left(b+c\right)\)

=>\(a^2=2\left(a+a\right)=4a\)

=>\(a^2-4a=0\)

=>a(a-4)=0

=>\(\left[{}\begin{matrix}a=0\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)

Do đó: a=b=c=4

\(T=4^3+4^{2023}+4=68+4^{2023}\)

Ai giúp mình với

a: Sửa đề: \(\dfrac{-2}{5}\cdot\dfrac{3}{4}+\dfrac{-2}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-2}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)

\(=-\dfrac{2}{5}\cdot\dfrac{4}{4}=-\dfrac{2}{5}\)

b: \(\left(\dfrac{5}{-3}-\dfrac{1}{2}\right)\cdot\dfrac{-6}{26}\)

\(=\left(\dfrac{-5}{3}-\dfrac{1}{2}\right)\cdot\dfrac{-3}{13}\)

\(=\dfrac{-13}{6}\cdot\dfrac{-3}{13}=\dfrac{3}{6}=\dfrac{1}{2}\)

c: \(\dfrac{-4}{15}:\left[-\dfrac{2}{3}+\dfrac{3}{5}\right]\)

\(=\dfrac{-4}{15}:\dfrac{-2\cdot5+3\cdot3}{15}\)

\(=-\dfrac{4}{15}\cdot\dfrac{15}{-1}=4\)

\(-\dfrac{2}{5}\cdot\dfrac{3}{4}+\dfrac{-2}{5}+\dfrac{1}{4}=-\dfrac{2}{5}\cdot\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{2}{5}\cdot\dfrac{4}{4}=-\dfrac{2}{5}\cdot1=-\dfrac{2}{5}\)

\(\left(\dfrac{5}{-3}-\dfrac{1}{2}\right)\cdot\left(-\dfrac{6}{26}\right)=\left(-\dfrac{10}{6}-\dfrac{3}{6}\right)\cdot\left(-\dfrac{6}{26}\right)=-\dfrac{13}{6}\cdot\left(-\dfrac{6}{26}\right)=\dfrac{1}{2}\)

\(-\dfrac{4}{15}:\left[\left(-\dfrac{2}{3}\right)+\dfrac{3}{5}\right]=-\dfrac{4}{15}:\left(-\dfrac{10}{15}+\dfrac{9}{15}\right)=-\dfrac{4}{15}:\left(-\dfrac{1}{15}\right)=-\dfrac{4}{15}\cdot\left(-15\right)=4\)

Đề không rõ ràng. Đề cần ghi rõ điều kiện, và yêu cầu đề. Bạn xem lại nhé.

\(\dfrac{1}{2\cdot2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}\)

\(=\dfrac{1}{4}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)