Bài 1:
Ta có:
VT=\(\left(a^2+b^2\right)\left(c^2+d^2\right)\)
=\(a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
=\(\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)
=\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\) = VP
Vậy đẳng thức được chứng minh
Bài 2:
a/P=\(x^2-2x+5\)
=\(\left(x^2-2x+1\right)+4\)
=\(\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow P\ge4\forall x\)
Vậy GTNN của P là 4 khi \(\left(x-1\right)^2=0\) hay x=1
b/Q=\(2x^2-6x\)
=\(2\left(x^2-3x\right)\)
=\(2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
=\(2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
\(\Rightarrow Q\ge-\dfrac{9}{2}\forall x\)
Vậy GTNN của Q là \(-\dfrac{9}{2}\) khi \(\left(x-\dfrac{3}{2}\right)^2=0\) hay \(x=\dfrac{3}{2}\)
c/\(M=x^2+y^2-x+6y+10\)
=\(x^2-x+\dfrac{1}{4}+y^2+6y+9+\dfrac{3}{4}\)
=\(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)
\(\Rightarrow M\ge\dfrac{3}{4}\forall x,y\)
Vậy GTNN của M là \(\dfrac{3}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\) và \(\left(y+3\right)^2=0\) hay \(x=\dfrac{1}{2}\) và y = -3
Bài 3:
a/Đặt A=\(x^2-6x+10\)
A=\(x^2-6x+9+1=\left(x-3\right)^2+1\)
Vì \(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+1\ge1>0\forall x\)
\(\Rightarrow A>0\forall x\)
\(\Rightarrow x^2-6x+10>0\forall x\)
b/Đặt B=\(4x-x^2-5\)
B=\(-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\)
Vì \(\left(x-2\right)^2\ge0\forall x\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1\le-1< 0\forall x\)
\(\Rightarrow B< 0\forall x\)
\(\Rightarrow4x-x^2-5< 0\forall x\)