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Мне прямо сейчас надо, на множители: а)100а^4-дробная черта наверху 1 внизу 9 b^2= б)9х^2-(х-1)^2= в)х^3+у^6.

Алгебра

Ответы

Захаров-Иванович

1)=100a^4 * 9b^2 -1 =(10a^2*3b-1)(10a^2*3b+1)=(30ab-1)(30ab+1)

2)=(3x-(x-+(x-1))=(3x-x+1)(3x+x-1)=(2x+1)(4x-1)

3)=x^3+(y^2)^3=(x+y^2)(x^2-xy^2+y^4)

ekasatkina
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krasa28vostok65
1. пусть скорость течения х. тогда скорость катера по течению 15+х, а против течения 15-х. тогда путь по течению занял 18/(15+х), а против течения 24/(15-х) 18/(15+х) + 24/(15-х)=3 сократим в 3 раза для легкости расчетов 6/(15+х) + 8/(15-х)=1 к одному знаменателю 6(15-х)/(15+х)(15-х) + 8(15+х)/(15-х)(15+х)=1 6(15-х) + 8(15+х)=(15-х)(15+х) 90-6х + 120+8х = 225-х² 210+2х = 225-х² х²+2х-15=0 d=2²+4*15=64 √d=8 x₁=(-2-8)/2=-5 отбрасываем отрицательное значение x₂=(-2+8)/2=3 км/ч ответ: скорость течения 3 км/ч 2. пусть скорость течения х. тогда скорость катера по течению 16+х, а против течения 16-х. тогда путь по течению занял 9/(16+х), а против течения 21/(16-х) 9/(16+х) + 21/(16-х)=2 к единому знаменателю 9(16-х)/(16+х)(16-х) + 21(16+х)/(16-х)(16+х)=2 9(16-х) + 21(16+х)=2(16²-х²) 144-9х+336+21х=512-2х² 144-9х+336+21х=512-2х² 480+12х=512-2х² 2х²+12х-32=0 х²+6х-16=0 d=6²+4*16=100 √d=10 x₁=(-6-10)/2=-8 отбрасываем отрицательное значение x₂=(-6+10)/2=2 км/ч ответ: скорость течения 2 км/ч

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Мне прямо сейчас надо, на множители: а)100а^4-дробная черта наверху 1 внизу 9 b^2= б)9х^2-(х-1)^2= в)х^3+у^6.
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