Предмет: Алгебра,
автор: frozenolaf1
решите пожалуйста номер 95,96,97
Приложения:
![](https://files.topotvet.com/i/5c2/5c26c6a95dbb18c4e0be393b7d4666ac.jpg)
Ответы
Автор ответа:
0
№ 95.
а) = 11 - 4^2 = 11- 16 = -5;
б)=(5sgrt2)^2 - 20^2 = 25*2 - 400 = 50 - 400 = - 350;
в) = (sgrt5)^2 - (2sgrt3)^2 = 5 - 4*3 = 5 - 12 = - 7;
г) = 1^2 - (3sgrt2)^2 = 1 - 9*2 = 1 - 18 = - 17.
96.а) (6- sgrt3)^2 = 6^2 - 2*6*sgrt3 + (sgrt)3^2 = 36 - 12 sgrt3 +3 = 39 - 12 sgrt3.
б) (sgrt5+sgrt7)^2 = 5 + 2*sgrt5*sgrt7 + 7 = 12 + 2sgrt35;
в) (2sgrt2 - sgrt8)^2 = (2 sgrt2)^2 - 2*2sgrt2*sgrt8 +(sgrt8)^2 = 4*2 - 4*sgrt16 +8=8 - 16 +8=0
97.
а) sgrt(6x+4)
x= 10; ⇒sgrt(6*10 +4) = sgrt 64 = 8;
x = 0; ⇒ sgrt(6*0 + 4) = sgrt 4 = 2
x =0,14; ⇒sgrt(6*0,14 +4) =sgrt 4,84 = 2,2
б) 3x^2 - 6;
x=sgrt7; ⇒ 3x^2 - 6 = 3*(sgrt7)^2 - 6 = 3*7 - 6 = 21 - 6 = 15;
x = - sgrt10; ⇒ 3x^2 - 6 = 3*(-sgrt10)^2 - 6 = 3*10 - 6 = 30 - 6 = 24;
x = 4sgrt3; ⇒ 3x^2 - 6 = 3*(4sgrt3)^2 - 6 = 3*16*3 - 6 = 144 - 6 = 138.
98. sgrt2*sgrt18 = sgrt(2*18) = sgrt36 = 6;
sgrt50*sgrt2= sgrt(50*2)= sgrt100 = 10;
sgrt45*sgrt80=sgrt(45*80)= sgrt 3600 = 60;
sgrt18 / sgrt2 = sgrt(18/2) = sgrt9 = 3;
sgrt50 / sgrt2 = sgrt(50 / 2) = sgrt25 = 5;
sgrt 540 / sgrt15 = sgrt(540 / 15) = sgrt36 = 6;
sgrt (25*9) =sgrt9 * sgrt25 = 3 * 5 = 15;
sgrt 7.1/9 = sgrt 64 / 9 = 8/3.
⇒
а) = 11 - 4^2 = 11- 16 = -5;
б)=(5sgrt2)^2 - 20^2 = 25*2 - 400 = 50 - 400 = - 350;
в) = (sgrt5)^2 - (2sgrt3)^2 = 5 - 4*3 = 5 - 12 = - 7;
г) = 1^2 - (3sgrt2)^2 = 1 - 9*2 = 1 - 18 = - 17.
96.а) (6- sgrt3)^2 = 6^2 - 2*6*sgrt3 + (sgrt)3^2 = 36 - 12 sgrt3 +3 = 39 - 12 sgrt3.
б) (sgrt5+sgrt7)^2 = 5 + 2*sgrt5*sgrt7 + 7 = 12 + 2sgrt35;
в) (2sgrt2 - sgrt8)^2 = (2 sgrt2)^2 - 2*2sgrt2*sgrt8 +(sgrt8)^2 = 4*2 - 4*sgrt16 +8=8 - 16 +8=0
97.
а) sgrt(6x+4)
x= 10; ⇒sgrt(6*10 +4) = sgrt 64 = 8;
x = 0; ⇒ sgrt(6*0 + 4) = sgrt 4 = 2
x =0,14; ⇒sgrt(6*0,14 +4) =sgrt 4,84 = 2,2
б) 3x^2 - 6;
x=sgrt7; ⇒ 3x^2 - 6 = 3*(sgrt7)^2 - 6 = 3*7 - 6 = 21 - 6 = 15;
x = - sgrt10; ⇒ 3x^2 - 6 = 3*(-sgrt10)^2 - 6 = 3*10 - 6 = 30 - 6 = 24;
x = 4sgrt3; ⇒ 3x^2 - 6 = 3*(4sgrt3)^2 - 6 = 3*16*3 - 6 = 144 - 6 = 138.
98. sgrt2*sgrt18 = sgrt(2*18) = sgrt36 = 6;
sgrt50*sgrt2= sgrt(50*2)= sgrt100 = 10;
sgrt45*sgrt80=sgrt(45*80)= sgrt 3600 = 60;
sgrt18 / sgrt2 = sgrt(18/2) = sgrt9 = 3;
sgrt50 / sgrt2 = sgrt(50 / 2) = sgrt25 = 5;
sgrt 540 / sgrt15 = sgrt(540 / 15) = sgrt36 = 6;
sgrt (25*9) =sgrt9 * sgrt25 = 3 * 5 = 15;
sgrt 7.1/9 = sgrt 64 / 9 = 8/3.
⇒
Автор ответа:
0
небольшая ошибка в 96 (а) там будет ответ 39-12 sgrt3
Автор ответа:
0
(6- sgrt3)^2 = 6^2 - 2*6*sgrt3 + (sgrt3)^2 = 36 - 12 sgrt3 +3 = 39 - 12 sgrt3.вот так
Автор ответа:
0
95
a)![( sqrt{11}+4)( sqrt{11}-4)=( sqrt{11}) ^{2}-4 ^{2}=11-16=-5 ( sqrt{11}+4)( sqrt{11}-4)=( sqrt{11}) ^{2}-4 ^{2}=11-16=-5](https://tex.z-dn.net/?f=%28+sqrt%7B11%7D%2B4%29%28+sqrt%7B11%7D-4%29%3D%28+sqrt%7B11%7D%29+%5E%7B2%7D-4+%5E%7B2%7D%3D11-16%3D-5+++++)
б)![(5 sqrt{2}-20)(5 sqrt{2} +20)=(5 sqrt{2}) ^{2}-20 ^{2}=50-400=-350 (5 sqrt{2}-20)(5 sqrt{2} +20)=(5 sqrt{2}) ^{2}-20 ^{2}=50-400=-350](https://tex.z-dn.net/?f=%285+sqrt%7B2%7D-20%29%285+sqrt%7B2%7D+%2B20%29%3D%285+sqrt%7B2%7D%29+%5E%7B2%7D-20+%5E%7B2%7D%3D50-400%3D-350++++)
в)![( sqrt{5}-2 sqrt{3})( sqrt{5}+2 sqrt{3})=( sqrt{5}) ^{2}-(2 sqrt{3}) ^{2}=5-12=-7 ( sqrt{5}-2 sqrt{3})( sqrt{5}+2 sqrt{3})=( sqrt{5}) ^{2}-(2 sqrt{3}) ^{2}=5-12=-7](https://tex.z-dn.net/?f=%28+sqrt%7B5%7D-2+sqrt%7B3%7D%29%28+sqrt%7B5%7D%2B2+sqrt%7B3%7D%29%3D%28+sqrt%7B5%7D%29+%5E%7B2%7D-%282+sqrt%7B3%7D%29+%5E%7B2%7D%3D5-12%3D-7++++++++)
г)![(3 sqrt{2}+1)(1-3 sqrt{2)} = (1+3 sqrt{2})(1-3 sqrt{2)} =1 ^{2}-(3 sqrt{2}) ^{2}=1-18= \ =-17 (3 sqrt{2}+1)(1-3 sqrt{2)} = (1+3 sqrt{2})(1-3 sqrt{2)} =1 ^{2}-(3 sqrt{2}) ^{2}=1-18= \ =-17](https://tex.z-dn.net/?f=%283+sqrt%7B2%7D%2B1%29%281-3+sqrt%7B2%29%7D+%3D+%281%2B3+sqrt%7B2%7D%29%281-3+sqrt%7B2%29%7D+%3D1+%5E%7B2%7D-%283+sqrt%7B2%7D%29+%5E%7B2%7D%3D1-18%3D+%5C+%3D-17++)
96
а)![(6- sqrt{3}) ^{2}=6 ^{2}-2cdot6cdot sqrt{3}+( sqrt{3}) ^{2}=36-12 sqrt{3}+3=39-12 sqrt{3} (6- sqrt{3}) ^{2}=6 ^{2}-2cdot6cdot sqrt{3}+( sqrt{3}) ^{2}=36-12 sqrt{3}+3=39-12 sqrt{3}](https://tex.z-dn.net/?f=%286-+sqrt%7B3%7D%29+%5E%7B2%7D%3D6+%5E%7B2%7D-2cdot6cdot+sqrt%7B3%7D%2B%28+sqrt%7B3%7D%29+%5E%7B2%7D%3D36-12+sqrt%7B3%7D%2B3%3D39-12+sqrt%7B3%7D+++++++)
б)![( sqrt{5}+ sqrt{7}) ^{2}=( sqrt{5}) ^{2}+2cdot sqrt{5}cdot sqrt{7}+( sqrt{7}) ^{2}= \ =5+2 sqrt{35}+7=12+2 sqrt{35} ( sqrt{5}+ sqrt{7}) ^{2}=( sqrt{5}) ^{2}+2cdot sqrt{5}cdot sqrt{7}+( sqrt{7}) ^{2}= \ =5+2 sqrt{35}+7=12+2 sqrt{35}](https://tex.z-dn.net/?f=%28+sqrt%7B5%7D%2B+sqrt%7B7%7D%29+%5E%7B2%7D%3D%28+sqrt%7B5%7D%29+%5E%7B2%7D%2B2cdot+sqrt%7B5%7Dcdot+sqrt%7B7%7D%2B%28+sqrt%7B7%7D%29+%5E%7B2%7D%3D+%5C+%3D5%2B2+sqrt%7B35%7D%2B7%3D12%2B2+sqrt%7B35%7D+++++++++++)
в)![(2 sqrt{2}- sqrt{8}) ^{2}=(2 sqrt{2}) ^{2}-2cdot2 sqrt{2} sqrt{8}+( sqrt{8}) ^{2}=8-16+8=0 (2 sqrt{2}- sqrt{8}) ^{2}=(2 sqrt{2}) ^{2}-2cdot2 sqrt{2} sqrt{8}+( sqrt{8}) ^{2}=8-16+8=0](https://tex.z-dn.net/?f=%282+sqrt%7B2%7D-+sqrt%7B8%7D%29+%5E%7B2%7D%3D%282+sqrt%7B2%7D%29+%5E%7B2%7D-2cdot2+sqrt%7B2%7D++sqrt%7B8%7D%2B%28+sqrt%7B8%7D%29+%5E%7B2%7D%3D8-16%2B8%3D0++++++++)
97
a)![sqrt{6x+4}= sqrt{6cdot10+4}= sqrt{64}=8, sqrt{6x+4}= sqrt{6cdot10+4}= sqrt{64}=8,](https://tex.z-dn.net/?f=+sqrt%7B6x%2B4%7D%3D+sqrt%7B6cdot10%2B4%7D%3D+sqrt%7B64%7D%3D8%2C)
![sqrt{6x+4}= sqrt{6cdot0+4}= sqrt{4}=2, sqrt{6x+4}= sqrt{6cdot0+4}= sqrt{4}=2,](https://tex.z-dn.net/?f=+sqrt%7B6x%2B4%7D%3D+sqrt%7B6cdot0%2B4%7D%3D+sqrt%7B4%7D%3D2%2C)
![sqrt{6x+4}= sqrt{6cdot0,14+4}= sqrt{0,84+4}= sqrt{4,84}=2,2 sqrt{6x+4}= sqrt{6cdot0,14+4}= sqrt{0,84+4}= sqrt{4,84}=2,2](https://tex.z-dn.net/?f=+sqrt%7B6x%2B4%7D%3D+sqrt%7B6cdot0%2C14%2B4%7D%3D+sqrt%7B0%2C84%2B4%7D%3D+sqrt%7B4%2C84%7D%3D2%2C2++)
б) 3х²-6
при х =√7
3·(√7)²-6=3·7-6=21-6=15
при х= -10
3·(-10)²-6=3·100-6=300-6=294
при х = 4√3
3·(4√3)²-6=3·16·3-6=144-6=138
a)
б)
в)
г)
96
а)
б)
в)
97
a)
б) 3х²-6
при х =√7
3·(√7)²-6=3·7-6=21-6=15
при х= -10
3·(-10)²-6=3·100-6=300-6=294
при х = 4√3
3·(4√3)²-6=3·16·3-6=144-6=138
Похожие вопросы
Предмет: Английский язык,
автор: LIANA1681
Предмет: Биология,
автор: vpvlchnk
Предмет: Химия,
автор: Vasya1122331
Предмет: Математика,
автор: shuka11
Предмет: Математика,
автор: Аноним