\(a,5,2x+7\dfrac{2}{5}=6\dfrac{3}{4}\\ \Rightarrow\dfrac{26}{5}x+\dfrac{37}{5}=\dfrac{27}{4}\\ \Rightarrow\dfrac{26}{5}x=-\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{8}\\ b,2,4:\left(\dfrac{-1}{2}-x\right)=1\dfrac{3}{5}\\ \Rightarrow\dfrac{12}{5}:\left(\dfrac{-1}{2}-x\right)=\dfrac{8}{5}\\ \Rightarrow\dfrac{-1}{2}-x=\dfrac{3}{2}\\ \Rightarrow x=-2\)

a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)

    \(\dfrac{3}{7}\)\(x\)         = - \(\dfrac{17}{35}\) + 0,4

     \(\dfrac{3}{7}\)\(x\)       = - \(\dfrac{3}{35}\)

        \(x\)       = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)

        \(x\)       = - \(\dfrac{1}{5}\)

b, 0,2.(\(x\) - 3) +2,4 = 10

    0,2.(\(x\) - 3)          = 10 - 2,4

    0,2.(\(x\) - 3)          = 7,6

           \(x\) - 3            = 7,6:0,2

           \(x\) - 3            = 38

            \(x\)                = 38 + 3

             \(x\)                = 41

\(\dfrac{3}{7}x-0,4=\dfrac{-17}{35}\)

\(\dfrac{3}{7}x=\dfrac{-17}{35}+0,4=\dfrac{-34}{70}+\dfrac{28}{70}\)

\(\dfrac{3}{7}x=\dfrac{-6}{70}=\dfrac{-3}{35}\)

\(x=\dfrac{-3}{35}:\dfrac{3}{7}=\dfrac{-3}{35}\cdot\dfrac{7}{3}\)

\(x=\dfrac{3\cdot\left(-1\right)\cdot7}{5\cdot7\cdot3}=-\dfrac{1}{5}\)

b) \(0,2\left(x-3\right)+2,4=10\)

\(0,2\left(x-3\right)=10-2,4=7,6\)

\(x-3=7,6:0,2=38\)

\(x=38+3=41\)

c) \(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\left(0,5-1\dfrac{3}{5}\right)\)

\(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\)

\(x:2,2=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\cdot\dfrac{6}{1}\)

\(x:2,2=\dfrac{-3\cdot\left(-11\right)\cdot2\cdot3}{2\cdot4\cdot10\cdot1}=\dfrac{99}{40}\)

\(x=\dfrac{99}{40}\cdot2,2=\dfrac{99}{40}\cdot\dfrac{22}{10}\)

\(x=\dfrac{99\cdot2\cdot11}{40\cdot2\cdot5}=\dfrac{1089}{200}\)

| x - 2,4| = \(\dfrac{1}{2}\)

\(\left[{}\begin{matrix}x-2,4=\dfrac{1}{2}(đk:x>2,4)\\x-2,4=-\dfrac{1}{2}(đk:x< 2,4)\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{2}+2,4\\x=-\dfrac{1}{2}+2,4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2,9(tm)\\x=1,9(tm)\end{matrix}\right.\)

vậy \(x\in\) { 1,9 ; 2,9}

x=2/9

I x - 2,4 I = 1/2

I x - 12/5 I = 1/2

I x I = 1/2 + 12/5

I x I =29/10

x = 29/10

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)

a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)

=>\(4⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

=>\(x\in\left\{16;4;25;1;49\right\}\)

b: 

Xét hpt \(\left\{{}\begin{matrix}\dfrac{x}{y}+2.\dfrac{y}{x}=3\left(1\right)\\2x^2-3y=-1\left(2\right)\end{matrix}\right.\) (đkxđ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\))

Từ (1) \(\Leftrightarrow\dfrac{x^2+2y^2}{xy}=3\Rightarrow x^2+2y^2=3xy\Leftrightarrow x^2-3xy+2y^2=0\)\(\Leftrightarrow x^2-xy-2xy+2y^2=0\Leftrightarrow x\left(x-y\right)-2y\left(x-y\right)=0\)\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x-2y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

Xét trường hợp \(x=y\), thay vào (2), ta có \(2x^2-3x=-1\Leftrightarrow2x^2-3x+1=0\) (3)

pt (3) có tổng các hệ số bằng 0 nên pt này có 2 nghiệm \(\left[{}\begin{matrix}x_1=1\\x_2=\dfrac{1}{2}\end{matrix}\right.\)(nhận)

Nếu \(x=1\Rightarrow y=1\) (vì \(x=y\)) (nhận)

Nếu \(x=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}\) (nhận)

Vậy ta tìm được 2 nghiệm của hpt đã cho là \(\left(1;1\right)\) và \(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\)

Xét trường hợp \(x=2y\), thay vào (2), ta có \(2.\left(2y\right)^2-3y=-1\Leftrightarrow8y^2-3y+1=0\) (4)

pt (4) có \(\Delta=\left(-3\right)^2-4.8.1=-23< 0\) nên pt này vô nghiệm.

Vậy hpt đã cho có tập nghiệm \(S=\left\{\left(1;1\right);\left(\dfrac{1}{2};\dfrac{1}{2}\right)\right\}\)

 Áp dụng BĐT Cauchy cho 3 số thực dương \(xy,yz,zx\), ta có \(xy+yz+zx\ge3\sqrt[3]{\left(xyz\right)^2}\). Do \(xy+yz+zx=3xyz\) nên\(3xyz\ge3\sqrt[3]{\left(xyz\right)^2}\) \(\Leftrightarrow3\sqrt[3]{\left(xyz\right)^2}\left(\sqrt[3]{xyz}-1\right)\ge0\) \(\Leftrightarrow\sqrt[3]{xyz}\ge1\) \(\Leftrightarrow xyz\ge1\)

ĐTXR \(\Leftrightarrow\left\{{}\begin{matrix}xy=yz=zx\\xy+yz+zx=3xyz\end{matrix}\right.\) \(\Leftrightarrow x=y=z=1\)

Ta có \(\dfrac{x}{1+y^2}=\dfrac{x\left(1+y^2\right)-xy^2}{1+y^2}=x-\dfrac{xy^2}{1+y^2}\ge x-\dfrac{xy^2}{2y}\)\(=x-\dfrac{xy}{2}\)

Tương tự, ta có \(\dfrac{y}{1+z^2}\ge y-\dfrac{yz}{2}\) và \(\dfrac{z}{1+x^2}\ge z-\dfrac{zx}{2}\). Từ đó suy ra \(\dfrac{x}{1+y^2}+\dfrac{y}{1+z^2}+\dfrac{z}{1+x^2}\ge x+y+z-\dfrac{xy+yz+zx}{2}\) \(=x+y+z-\dfrac{3}{2}xyz\) . Từ đây suy ra \(Q\ge x+y+z\ge\sqrt[3]{xyz}\ge1\). ĐTXR \(\Leftrightarrow x=y=z=1\). 

Vậy GTNN của \(Q\) là \(1\) đạt được khi \(x=y=z=1\)

 Dạ thưa thầy, chỗ kia con sửa là \(Q\ge x+y+z\ge3\sqrt[3]{xyz}\ge3\) ạ. GTNN của Q là 3 khi \(x=y=z=1\)

\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)

\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}=\dfrac{7}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}+\dfrac{7}{2}x=-\dfrac{1}{3}\\ \Rightarrow\left(\dfrac{1}{2}+\dfrac{7}{2}\right)x+\dfrac{5}{2}=-\dfrac{1}{3}\\ \Rightarrow4x=-\dfrac{17}{6}\\ \Rightarrow x=-\dfrac{17}{24}.\)

\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{2}x-3\dfrac{1}{2}x=-\dfrac{1}{3}-2\dfrac{1}{2}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{7}{2}\right)x=-\dfrac{1}{3}-\dfrac{5}{2}\\ \Rightarrow\dfrac{-6}{2}x=-\dfrac{17}{6}\\ \Rightarrow-3x=-\dfrac{17}{6}\\ \Rightarrow x=\left(-\dfrac{17}{6}\right):\left(-3\right)\\ \Rightarrow x=\dfrac{17}{18}\)