Question

СОЧ помогите пожалуйста ​

Answer 1

Ответ:

$(5+b)(5-b)-b ^ { 2 } = 25-b^2-b^2 =25-2b^2\\ (x+5)6+(x-3) ^ { 2 } = 6x+30+x^2-6x+9 =39+x^{2} \\ (9-b)(9+b)+(3-b) ^ { 2 } = 81-b^2+9-6b+b^2 =90-6b\\ (2a-11)(11+2a)-(2a-5) ^ { 2 } =0\\ 4a^2-121-4a^2+20a-25 = 0 \\-121+20a-25=0\\20a=146\\\\ a= \dfrac{ 146 }{ 20 } = 7 \dfrac { 3 } { 10 }$

$( \dfrac{ 1 }{ 5 } a ^ { 3 } +0.3) ^ { 3 } = \dfrac{ 1 }{ 125 } a ^ { 9 } + \dfrac{ 9 }{ 250 } a ^ { 6 } + \dfrac{ 27 }{ 500 } a ^ { 3 } +0.027$

${ \left( { a }^{ 3 } + \dfrac{ 1 }{ 3 } b ^ { 2 } \right) }^{ 3 } = \dfrac{ a ^ { 3 } b ^ { 4 } }{ 3 } +b ^ { 2 } a ^ { 6 } +a ^ { 9 } + \dfrac{ b ^ { 6 } }{ 27 }$

$25-(13+t) ^ { 2 } =0 \\ 25-\left( 169+26t+t^2 \right) = 0 \\25-169-26t-t^2 = 0\\-t^2-26t-144=0\\\sqrt{D} =\sqrt{676-4*(-1)(-144)} =10\\ x= \dfrac{ 26+10 }{ -2 } =-18\\ x= \dfrac{ 26-10 }{ -2 } =-8$

$100-(z-19) ^ { 2 } =0 \\ -z^2+38z-261 = 0 \\\sqrt{D} =\sqrt{1444-4*(-1)(-261)} =20\\ x= \dfrac{ -38+20 }{ -2 } =9\\ x= \dfrac{ -38-20 }{ -2 } =29$