\(VT=a^2+4b^2+1-4ab+2a-4b+b^2-2b+1+1\)

\(VT=\left(a-2b+1\right)^2+\left(b-1\right)^2+1>0\) (đpcm)

\(a^2+5b^2-4ab+2a-6b+3\)

\(=a^2-4ab+2a+5b^2-6b+3\)

\(=a^2-2a\left(2b-1\right)+5b^2-6b+3\)

\(=a^2-2.a.\frac{2b-1}{2}+\left(\frac{2b-1}{2}\right)^2+5b^2-6b-\left(\frac{2b-1}{2}\right)^2+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-\frac{\left(2b-1\right)^2}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-\frac{4b^2-4b+1}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-b^2+b-\frac{1}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+4b^2-5b+\frac{11}{4}\)

\(=\left(a-\frac{2b-1}{2}\right)^2+\left(2b\right)^2-2.2b.\frac{5}{4}+\frac{25}{16}+\frac{19}{16}\)

\(=\left(a-\frac{2b-1}{2}\right)^2+\left(2b-\frac{5}{4}\right)^2+\frac{19}{16}\)

Vì \(\left(a-\frac{2b-1}{2}\right)^2\ge0;\left(2b-\frac{5}{4}\right)^2\ge0=>\left(a-\frac{2b-1}{2}\right)^2+\left(2b-\frac{5}{4}\right)^2+\frac{19}{16}\ge\frac{19}{16}>0\) (với mọi a,b)  (đpcm)

\(a^2+5b^2-4ab+2a-6b+3\)

\(=\left(a^2-4ab+4b^2\right)+\left(2a-4b\right)+1+\left(b^2-2b+1\right)+1\)

\(=\left(a-2b\right)^2+2\left(a-2b\right)+1+\left(b^2-2b+1\right)+1\)

\(=\left(a-2b+1\right)^2+\left(b-1\right)^2+1\ge1\forall a;b\)

Mà \(1>0\) nên \(a^2+5b^2-4ab+2a-6b+3>0\forall a;b\)(đpcm)

a, \(\left(a^2+b^2-2ab+2a-2b+1\right)+\left(b^2-2b+1\right)=0\)

=> \(\left(a-b+1\right)^2+\left(b-1\right)^2=0\)

Mà \(\left(a-b+1\right)^2\ge0,\left(b-1\right)^2\ge0\)

=> \(\hept{\begin{cases}a-b+1=0\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}}\)

b,Tương tự 

\(\left(a-2b+1\right)^2+\left(b-1\right)^2=0\)

=>\(\hept{\begin{cases}a=1\\b=1\end{cases}}\)

`a^2+4ab-5b^2=0`

`<=>a^2+4ab+4b^2-9b^2=0`

`<=>(a+2b)^2-9b^2=0`

`<=>(a+2b-3b)(a+2b+3b)=0`

`<=>(a-b)(a+5b)=0`

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-5b\end{matrix}\right.\)

`Q={2a-b}/{a-b}+{3a-2b}/{a+b}`

Với `a=b` `=>` giá trị vô nghĩa

Với `a=-5b` 

`Q={-10b-b}/{-5b-b}+{-15b-2b}/{-5b+b}`

`Q={-11b}/{-6b}+{-17b}/{-4b}`

`Q=11/6+17/4`

`Q=73/12`

\(5a^2+10b^2-6ab-4a+2b+3\)

\(=\left(a^2-6ab+9b^2\right)+\left(4a^2-4a+1\right)+\left(b^2+2b+1\right)+1\)

\(=\left(a-3b\right)^2+\left(2a-1\right)^2+\left(b+1\right)^2+1>0\left(đpcm\right)\)

câu b đề sai ak

1) ta có: a(b^2 -1)(c^2 -1)+b(a^2 -1)(c^2 -1)+c(a^2-1)(b^2-1)

=(ab^2 -a)(c^2-1)+(ba^2 -b)(c^2-1)+(ca^2-c)(b^2-1)

 đén đây nhân bung ra hết rồi rút gọn và thay a+b+c=abc là đc

a:Sửa đề:  \(a^2-4ab+4b^2\)

\(=a^2-2\cdot a\cdot2b+4b^2\)

\(=\left(a-2b\right)^2\ge0\)(luôn đúng)

b: \(-2a^2+a-1\)

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

\(=-2\left(a^2-2\cdot a\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{7}{16}\right)\)

\(=-2\left(a-\dfrac{1}{2}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}< 0\forall x\)