sao giống câu hỏi của mình thế chỉ khác số bạn biết làm ko chỉ mình đi

\(2D=x^2-4xy+4y^2+x^2-12x+36+6y^2-36y+54+10\)\(2D=\left(x-2y\right)^2+\left(x-6\right)^2+6\left(y-3\right)^2+10\)

\(2D\ge10\) => D>=5 khi x=2y=6

\(F=3x^2+x+4=3\left(x^2+\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{47}{12}\)

F=\(3\left(x+\dfrac{1}{6}\right)^2+\dfrac{47}{12}\ge\dfrac{47}{12}\) khi x=-1/6

\(2E=4x^2-4xy+y^2+y^2-4y+4+3996\)

\(2E=\left(2x-y\right)^2+\left(y-2\right)^2+3996\ge3996\)

E>=1998 khi 2x=y=2

bài 4;

\(B=-3x^2+x=-3\left(x^2-\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{1}{12}\)

\(B=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}\le\dfrac{1}{12}\)

khi x=1/6

bài 5:

\(a,\left(x+2\right)^2=0=>x=-2\)

\(b,\left(x-6\right)^2+\left(y+1\right)^2=0\rightarrow\left\{{}\begin{matrix}x=6\\y=-1\end{matrix}\right.\)

c,\(x^2+2y^2-2xy-2x+2=0\)

\(x^2-4xy+4y^2+x^2-4x+4=0\)

\(\left(x-2y\right)^2+\left(x-2\right)^2=0\rightarrow\left\{{}\begin{matrix}x=2y\\x=2\end{matrix}\right.\)

đây nhá bạn, khá tốn time của mình

\(\left\{{}\begin{matrix}5x=5m\\y=2x-m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=10-m+1=11-m\end{matrix}\right.\)

Thay vào ta đc 

\(2m^2-3\left(11-m\right)=2\Leftrightarrow2m^2-33+3m=2\Leftrightarrow2m^2+3m-35=0\Leftrightarrow m=\dfrac{7}{2};m=-5\)

\(a)\)

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

\(\Leftrightarrow x-x^2+1=3x+1\)

\(\Leftrightarrow x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

a: =>(x-1)(x-2)=0

=>x=1 hoặc x=2

b: TH1: x>=0

=>2x=3x+2

=>x=-2(loại)

TH2: x<0

=>-2x=3x+2

=>-5x=2

=>x=-2/5(nhận)

c: TH1: x>=0

=>2x=3x+4

=>-x=4

=>x=-4(loại)

TH2: x<0

=>-2x=3x+4

=>-5x=4

=>x=-4/5(nhận)

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

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

\(x+\frac{4}{x^2}-3x+2-x+\frac{1}{x^2}-4x+3=2x+\frac{5}{x^3}-4x+3\\ \Leftrightarrow-2x+\frac{5}{x^2}+2=2x+\frac{5}{x^3}\\ \Leftrightarrow4x=-2\\ \Leftrightarrow x=-\frac{1}{2}\)