Question

Помогите с решением ! 30 балов

Answer 1

Объяснение:

$1)\; \; (4x+1)^2+(3-2x)(8x+1)=7\\\\16x^2+8x+1+24x+3-16x^2-2x=7\\\\30x=3\\\\x=0,1\\\\\\2)\; \; (x^4-3)(x^4+3)-(x^4-5)^2=x^8-9-(x^8-10x^4+25)=10x^4-34=2(5x^4-17)\\\\\\3)\; \; 25x^{10}+y^8-10x^5y^4=(5x^5)^2-2\cdot (5x^5)\cdot y^4+(y^4)^2=(5x^5-y^4)^2\\\\(5x-4)^2-(x+7)^2=(5x-4-x-7)(5x-4+x+7)=(4x-11)(6x+3)\\\\216+b^3c^3=6^3+(bc)^3=(6+bc)(36-6bc+b^2c^2)$

Answer 2

Объяснение:

$1)(4x + 1) {}^{2} + (3 - 2x) \times (8x + 1) = 7 \\ \\ 16x {}^{2} + 8x + 1 + 24x + 3 - 16x {}^{2} - 2x = 7 \\ \\ 30x + 4 = 7 \\ \\ 30x = 7 - 4 \\ \\ 30x = 3 \\ \\ x = \frac{3}{30} = \frac{1}{10} \\ \\ \\ 2)(x {}^{4} - 3) \times (x {}^{4} + 3) - (x {}^{4} - 5) \\ \\ x {}^{8} - 9 - (x {}^{8} - 10x {}^{4} + 25) \\ \\ x {}^{8} - 9 - x {}^{8} + 10x {}^{4} - 25 \\ \\ - 34 + 10x {}^{4} \\ \\ \\ \\ 3) \\ 1)25x {}^{10} + y {}^{8} - 10x {}^{5} y {}^{4} \\ \\ 25x {}^{10} - 10x {}^{5} y {}^{4} + y {}^{8} \\ \\ (5x {}^{5} - y {}^{4} ) {}^{2} \\ \\ 2)(5x - 4) {}^{2} - (x + 7) {}^{2} \\ \\ (5x - 4 - (x + 7)) \times (5x - 4 + (x + 7)) \\ \\ (5x - 4 - x - 7) \times (5x - 4 + x + 7) \\ \\ (4x - 11) \times (6x + 3) \\ \\ (4x - 11) \times 3 \times (2x + 1) \\ \\ 3 \times (4x - 11) \times (2x + 1) \\ \\ \\ \\ 3)216 + b {}^{3} c {}^{3} \\ \\ 6 {}^{3} + (bc) {}^{3} \\ \\ (6 + bc) \times (6 {}^{2} - 6bc + (bc) {}^{2} \\ \\ (6 + bc) \times (36 - 6bc + b {}^{2} c {}^{2} )$

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