bài 5 :
+) ta có : \(A=x^2-4x+18=x^2-4x+4+14\)
\(=\left(x-2\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(B=x^2-x+2=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(C=x^2+2y^2-2xy-2y+15=x^2-2xy+y^2+y^2-2y+1+14\)
\(=\left(x-y\right)^2+\left(y-1\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
bài 6 :
+) ta có : \(M=x^2-10x+3=x^2-10x+25-22=\left(x-5\right)^2-22\ge-22\)
\(\Rightarrow M_{min}=-22\) khi \(x=5\)
+) ta có : \(N=x^2+6x-5=x^2+6x+9-14=\left(x+3\right)^2-14\ge-14\)
\(\Rightarrow N_{min}=-14\) khi \(x=-3\)
+) ta có : \(P=x^2+y^2-4x+20=x^2-4x+4+y^2+16=\left(x-2\right)^2+y^2+16\ge16\)
\(\Rightarrow P_{min}=16\) khi \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
+) ta có : \(Q=x\left(x-3\right)=x^2-3x=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)
\(\Rightarrow Q_{min}=\dfrac{-9}{4}\) khi \(x=\dfrac{3}{2}\)
bài 7 :
+) ta có : \(A=-x^2-12x+3=-\left(x^2+12x+36\right)+39=-\left(x+6\right)^2+39\le39\)
\(\Rightarrow A_{max}=39\) khi \(x=-6\)
+) ta có : \(B=-4x^2+4x+7=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\le11\)
\(\Rightarrow B_{max}=11\) khi \(x=2\)
bài 8 :
a) ta có : \(16x^2-9=0\Leftrightarrow x^2=\dfrac{9}{16}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
b) ta có : \(\left(x-2\right)^2-x^2=4\Leftrightarrow x^2-4x+4-x^2-4=0\Leftrightarrow x=0\)
c) ta có : \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=\dfrac{-255}{2}\)
d) ta có : \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)=16\)
\(\Leftrightarrow4x^2-12x+9-4x^2+1-16=0\Leftrightarrow-12x-6=0\Leftrightarrow x=\dfrac{-1}{2}\)
e) ta có : \(\left(x-2\right)\left(x+2\right)-x\left(x-2\right)=1\)
\(\Leftrightarrow x^2-4-x^2+2x-1=0\Leftrightarrow x=\dfrac{5}{2}\)
