\(\text{a) }2x^2+y^2-2xy-2x+3\\ =\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\\ =\left(x-y\right)^2+\left(x-1\right)^2+2\)

\(\text{Ta có : }\left(x-y\right)^2\ge0\\ \left(x-1\right)^2\ge0\\ \Rightarrow\left(x-y\right)^2+\left(x-1\right)^2\ge0\\ \Rightarrow\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\) \(\text{Dấu }"="\text{xảy ra khi : }\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1-y=0\\x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

Vậy giá trị nhỏ nhất biểu thức \(=2\) khi \(x=1\) và \(y=1\)

Điên à. Mẫu thức luôn khác không mà

\(\dfrac{5}{8}-\dfrac{3}{8}:\left(-\dfrac{3}{4}\right)-\dfrac{1}{4}\)

\(=\dfrac{5}{8}-\dfrac{3}{8}\cdot\left(-\dfrac{3}{4}\right)-\dfrac{1}{4}\)

\(=\dfrac{5}{8}+2-\dfrac{1}{4}\)

\(=\dfrac{19}{8}\)

\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)

\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)

\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)

\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)

\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)

\(=\dfrac{123}{456}\left(1-2\right)\)

\(=-\dfrac{123}{456}\)

\(A=x^2-2x+y^2-4y+6\\ A=x^2-2x+y^2+-4y+1+4+1\\ A=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\\ A=\left(x-1\right)^2+\left(y-2\right)^2+1\)

\(\text{Ta có : }\left(x-1\right)^2\ge0\\ \left(y-2\right)^2\ge0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\\ \Leftrightarrow A=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\)

Dấu \("="\) xảy ra khi :\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(A_{\left(Min\right)}=1\) khi \(x=1\) và \(y=2\)

\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)

\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)

\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)

\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)

\(x^3-25x=0\\\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x^2-5^2\right)=0\\ \Leftrightarrow x\left(x+5\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right. \)

Vậy \(x=0\) hoặc \(x=-5\) hoặc \(x=5\)

\(\text{a) }2x\left(12x-5\right)-8x\left(3x-1\right)=30\\ \Leftrightarrow24x^2-10x-24x^2+8x=30\\ \Leftrightarrow\left(24x^2-24x^2\right)-\left(10x-8x\right)=24\\ \Leftrightarrow-2x=24\\ \Leftrightarrow x=-12\\ \text{Vậy }x=-12\\ \\ \)

\(\text{b) }3x\left(3-2x\right)+6x\left(x-1\right)=15\\ \Leftrightarrow9x-6x^2+6x^2-6x=15\\ \Leftrightarrow\left(9x-6x\right)-\left(6x^2-6x^2\right)=15\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\)