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

\(\left(2a-b\right)^2-b^2+\left(2b-a\right)^2-a^2=0\)

\(\left(2a-b-b\right)\left(2a-b+b\right)+\left(2b-a-a\right)\left(2b-a+a\right)=0\)

\(2a\left(a-b\right)-2b\left(a-b\right)=0\)

\(2\left(a-b\right)^2=0\)

a-b=0

a=b

\(\sqrt{2x-3}=x-3\left(đkxđ:x\ge\dfrac{3}{2}\right)\)

\(\Rightarrow2x-3=\left(x-3\right)^2\)             (\(x\ge3\))

\(\Leftrightarrow2x-3=x^2-6x+9\)

\(\Leftrightarrow x^2-8x+12=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=6\end{matrix}\right.\)

Vậy....

a)\(\dfrac{2x-1}{x^2-1}=0\)    (\(ĐKXĐ:x\ne\pm1\))

\(\Rightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

b)\(\dfrac{\left(x+2\right)\left(x-3\right)}{2x-5}=0\)      (\(ĐKXĐ:x\ne\dfrac{5}{2}\))

\(\Rightarrow\left(x+2\right)\left(x-3\right)=0\)

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

c)\(\dfrac{x^2+x}{2x+1}=0\)        (\(ĐKXĐ:x\ne-\dfrac{1}{2}\))

\(\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(2xy+16-x^2-y^2\)

\(=16-\left(x^2-2xy+y^2\right)\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

Để A>1 thì 

\(\dfrac{x+5}{x+8}>1\)

\(\dfrac{x+5}{x+8}-1>0\)

\(\dfrac{x+5-x-8}{x+8}>0\)

\(\dfrac{-3}{x+8}>0\) 

Vì -3<0 nên

\(x+8< 0\)

\(x< 8\)

Vậy.....

\(C=\left(2-x\right)\left(x+4\right)\)

    \(=2x+8-x^2-4x\)

    \(=-x^2-2x+8\)

    \(=9-\left(x+1\right)^2\le9\)

Dấu = xảy ra khi x+1=0  \(\Leftrightarrow x=-1\)

Vậy \(Max_C=9\) tại x=-1

\(\left\{{}\begin{matrix}a-b=\sqrt{2}+1\\a-c=\sqrt{2}-1\end{matrix}\right.\)

=> \(a-b+b-c=\sqrt{2}+1+\sqrt{2}-1\)

\(\Leftrightarrow a-c=2\sqrt{2}\)

\(B=a^2+b^2+c^2-ab-bc-ac\)

\(=\dfrac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2}\)

\(=\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2}{2}\)

\(=\dfrac{3+2\sqrt{2}+3-2\sqrt{2}+8}{2}\)

\(=\dfrac{6+8}{2}=7\)

\(\left(2x-1\right)^2-4x^2+3=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+3=0\)

\(\Leftrightarrow-4x+4=0\)

\(\Leftrightarrow-4x=-4\)

=> x=1

Vậy....

\(\left(x-2\right)^2-9=0\)

\(\Leftrightarrow\left(x-2+3\right)\left(x-2-3\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)

Vậy....

\(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3\)=0

\(\Leftrightarrow x=2\)

Có: \(a:7\) dư 4

=> \(a=7k+4\)

\(b:7\) dư 3

=> \(b=7k+3\)

Ta có: \(a.b=\left(7k+4\right)\left(7k+3\right)\)

                 \(=49k^2+49k+12\)

                  \(=49k^2+49k+7+5\)

Vì \(49k^2⋮7\) , \(49k⋮7\) , \(7⋮7\)

=> a.b : 7 dư 5

\(\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}\)

\(=\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}\)

\(=\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)

\(=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}-\sqrt{5}+1=1\)

b)\(\sqrt{33+8\sqrt{2}-\sqrt{33-8\sqrt{2}}}\)

​​\(=\sqrt{33+8\sqrt{2}-\sqrt{\left(\sqrt{32}-1\right)^2}}\)

\(=\sqrt{33+8\sqrt{2}-\sqrt{32}+1}\)

\(=\sqrt{34+4\sqrt{2}}\)