ĐKXĐ: \(x\ne2;y\ne-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{x-2}+\frac{2}{y+1}=\frac{17}{5}\\\frac{2\left(x-2\right)+2}{x-2}+\frac{y+1+1}{y+1}=\frac{26}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{x-2}+\frac{2}{y+1}=\frac{17}{5}\\2+\frac{2}{x-2}+\frac{1}{y+1}=\frac{26}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{x-2}+\frac{2}{y+1}=\frac{17}{5}\\\frac{2}{x-2}+\frac{1}{y+1}=\frac{11}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x-2}=1\\\frac{1}{y+1}=\frac{1}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Tất cả đều trừ 1 vào mỗi thừa số=>tử số là:x-29(cái cuối+3)

=> cả cái đó=0

=> x-29=0 

=>x=29

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Rightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)+\left(\frac{x-44}{5}+3\right)=0\)

\(\Rightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}+\frac{x-29}{5}=0\)

\(\Rightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

\(\Rightarrow x-29=0\)

\(\Rightarrow x=29\)

Câu 1 mình nghĩ nó khá đơn giản rồi, bạn tính ra ngay thôi

Câu 2: Mình nghĩ là tìm min chứ ko phải max

Vì \(\left(-\frac{2}{3}+\frac{1}{2}x\right)^2\ge0\Rightarrow A=\left(-\frac{2}{3}+\frac{1}{2}x\right)^2-2,5\ge2,5\)

\(\Rightarrow A_{min}=2,5\Leftrightarrow\left(-\frac{2}{3}+\frac{1}{2}x\right)^2=0\Leftrightarrow-\frac{2}{3}+\frac{1}{2}x=0\Leftrightarrow\frac{1}{2}x=\frac{2}{3}\Leftrightarrow x=\frac{4}{3}\)

A đạt giá trị nhỏ nhất là 2,5 khi x=4/3

Câu 3: 

\(x=\frac{26}{7+b}\) âm khi 7+b âm <=> 7+b<0 <=> b<-7

vì b là số nguyên lớn nhất nên b=-8

\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)

TH1 : \(\Rightarrow x+\frac{1}{2}=0\)

\(x=0-\frac{1}{2}\)

\(x=\frac{-1}{2}\)

TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)

\(2x=0+\frac{2}{3}=\frac{2}{3}\)

\(x=\frac{2}{3}\div2=\frac{1}{3}\)

\(\Rightarrow x=\frac{-1}{2};\frac{1}{3}\)

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

\(x=\frac{3}{5}^2-\frac{1}{5}^2\)

\(x=\frac{2}{5}\)

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

Xét phương trình: \(\frac{2x}{3}+\frac{2x-1}{5}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}+\frac{2x-1}{5}=4\)

\(\Leftrightarrow x+\frac{2x-1}{5}=4\Leftrightarrow\frac{5x+2x-1}{5}=4\)

\(\Leftrightarrow7x-1=20\Leftrightarrow x=3\)

Để hai phương trình \(\frac{2x}{3}+\frac{2x-1}{5}=4-\frac{x}{3}\)và \(\left(k+1\right)x+k=26\)tương đương thì:

x = 3 là nghiệm của \(\left(k+1\right)x+k=26\)

\(\Rightarrow3\left(k+1\right)+k=26\Leftrightarrow3k+3+k=26\)

\(\Leftrightarrow4k=23\Leftrightarrow k=\frac{23}{4}\)

Vậy \(k=\frac{23}{4}\)thì hai phương trình trên tương đương