rozwiąż równaniaa)2x(x+4)-(3x-1)²=6-7*(x²-1)b)(x+2)*(x-3)-4=(x+1)²+1c)x*(x-3)*(x+1)-x³=2-2x²d)2*[(x-.... Question from @dami554

dami554 @dami554

November 2018 1 7 Report

rozwiąż równania

a)2x(x+4)-(3x-1)²=6-7*(x²-1)

b)(x+2)*(x-3)-4=(x+1)²+1

c)x*(x-3)*(x+1)-x³=2-2x²

d)2*[(x-4)²-2*(x+1)²+(4x-2)*(4x+2)]-28x²=19-27x

e)4*(x-3)-(x+1)*(x-1)=5*(x-1)-(x+3)²-2

niech ktoś mi to wytłumaczy bo nieczaje

za pierwszą odp daje naj ;D

krysta $a)\\2x\cdot (x+4)-(3x-1)^2=6-7 \cdot (x^2-1)\\ \\2x^{2}+8x-(9x^{2}-6x+1)=6-7x^2+7\\ \\2x^{2}+8x-9x^{2}+6x-1=6-7x^2+7$

$2x^{2}+8x-9x^{2}+6x+7x^{2}=6+7+1\\ \\14x=14/:14\\ \\x=1$

$b)\\(x+2)\cdot (x-3)-4=(x+1)^2+1\\ \\x^{2}-3x+2x-6-4=x^{2}+2x+1+1\\ \\x^{2}-3x+2x-x^{2}-2x = 1+1+6+4\\ \\-3x=12/:(-3)\\ \\x=-4$

$c)\\x\cdot (x-3)\cdot (x+1)-x^3=2-2x^2\\ \\x(x^{2}+x-3x-3)-x^{3}+2x^{2}=2\\ \\x^{3}+x^{2}-3x^{2}-3x-x^{3}+2x^{2}=2\\ \\-3x=2/:(-3)\\\\x=-\frac{2}{3}$

$d)\\2\cdot [(x-4)^2-2(x+1)^2+(4x-2)\cdot (4x+2)]-28x^2=19-27x\\ \\2\cdot [x^{2}-8x+16-2(x^{2}+2x+1)+16x^{2}-4]-28x^2=19-27x\\ \\2\cdot [x^{2}-8x+16-2x^{2}-4x-2+16x^{2}-4]-28x^2=19-27x\\ \\2x^{2}-16x+32-4x^{2}-8x-4+32x^{2}-8-28x^2+27x-19=0\\ \\2x^{2}-3x+1=0$

$\Delta =b^{^{2}}-4ac=(-3^{2})-4\cdot 2\cdot 1=9-8=1\\ \\\sqrt{\Delta }=\sqrt{1}=1\\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{3-1}{2\cdot 2}=\frac{2}{4}=\frac{1}{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{3+1}{2\cdot 2}=\frac{4}{4}=1$

$e)\\4\cdot (x-3)-(x+1)\cdot (x-1)=5\cdot (x-1)-(x+3)^2-2\\ \\4x-12-(x^{2}-1)=5x-5-(x^{2}+6x+9)-2\\ \\4x-12-x^{2}+1=5x-5-x^{2}-6x-9-2\\ \\4x-x^{2}-5x+x^{2}+6x=-5-9-2+12-1\\ \\5x=-5/:5\\ \\x=-1$

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