解: $x²-6x+9=16$ $(x-3)²=16$ $x-3=±4$ ${x}_{1}=7,{x}_{2}=-1$
解: $x²+10x+25=16$ $(x+5)²=16$ $x+5=±4$ ${x}_{1}=-1,{x}_{2}=-9$ 解: $x²+2x+1=8$ $(x+1)²=8$ $x+1=±2\sqrt{2}$ ${x}_{1}=-1+2\sqrt{2},{x}_{2}=-1-2\sqrt{2}$
解: $x²-6x+9=10$ $(x-3)²=10$ $x-3=±\sqrt{10}$ ${x}_{1}=3+\sqrt{10},{x}_{2}=3-\sqrt{10}$ 解: $t²-t+\frac{1}{4}=\frac{9}{4}$ $(t-\frac{1}{2})²=\frac{9}{4}$ $t-\frac{1}{2}=±\frac{3}{2}$ ${t}_{1}=2,{t}_{2}=-1$
解: $x²+3x+\frac{9}{4}=\frac{21}{4}$ $(x+\frac{3}{2})²=\frac{21}{4}$ $x+\frac{3}{2}=±\frac{\sqrt{21}}{2}$ ${x}_{1}=\frac{-3+\sqrt{21}}{2},{x}_{2}=\frac{-3-\sqrt{21}}{2}$
解:因为 $a=2,b=-5,c=0$ $所以 b²-4ac$ $=(-5)²-4×2×0=25＞0$ x= $\frac{-b±\sqrt{b²-4ac} }{2a}$ = $\frac{-(-5)±\sqrt{25}}{2×2}$ = $\frac{5±5}{4}$ ${x}_{1}=\frac{5}{2},{x}_{2}=0$
解:因为 $a=2,b=3,c=1$ 所以 $b²-4ac$ = $3²-4×2×1=1$ x= $\frac{-b±\sqrt{b²-4ac} }{2a}$ = $\frac{-3±1}{2×2}$ ${x}_{1}=-\frac{1}{2},{x}_{2}=-1$ 解: $a=3,b=-1,c=-2$ $b²-4ac$ $=(-1)²-4×3×(-2)=25$ x= $\frac{-b±\sqrt{b²-4ac} }{2a}$ = $\frac{1±\sqrt{25}}{2×3}$ ${x}_{1}=1,{x}_{2}=-\frac{2}{3}$ 解:因为 $a=4,b=-2,c=-1$ $所以 b²-4ac$ $=(-2)²-4×4×(-1)=20$ x= $\frac{-b±\sqrt{b²-4ac} }{2a}$ = $\frac{2±\sqrt{20}}{2×4}$ = $\frac{1±\sqrt{5}}{4}$ ${x}_{1}=\frac{1+\sqrt{5}}{4},{x}_{2}=\frac{1-\sqrt{5}}{4}$
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