Question

упростите и найдите значение 4x^2+289+68x где x=0, x=8.5
25y^2-70y+49 где y= 1/5, y=-2

Answer 1

$\displaystyle\bf\\1)\\\\4x^{2} +289+68x=4x^{2} +68x+289=(2x)^{2} +2\cdot 2x\cdot 17+17^{2} =\\\\=(2x+17)^{2} \\\\x=0\\\\(2x+17)^{2} =(2\cdot 0+17)^{2} =17^{2} =289\\\\x=8,5\\\\(2x+17)^{2} =(2\cdot 8,5+17)^{2} =(17+17)^{2} =34^{2} =1156\\\\\\2)\\\\25y^{2} -70y+49=(5y)^{2} -2\cdot 5y\cdot 7+7^{2} =(5y-7)^{2} \\\\y=\frac{1}{5} \\\\\Big(5\cdot\frac{1}{5} -7\Big)^{2} =\Big(1-7\Big)^{2} =\Big(-6\Big)^{2} =36$

$\displaystyle\bf\\y=-2\\\\(5y-7)^{2} =\Big[5\cdot(-2)-7\Big]^{2} =\Big(-10-7\Big)^{2} =\Big(-17\Big)^{2} =289$