Question

помогите пж алгебра9 класс​

Answer 1

Ответ:

Объяснение:

1)$10x^3-20x^2-x+2=0$

Решим группировкой

$10x^2(x-2)-(x-2)=0\\(x-2)(10x^2-1)=0\\x=2\\10x^2=1\\x^2=0.1\\x_{1}=\sqrt{0.1}\\x_{2}=-\sqrt{0.1}$

ответ: 2; $\sqrt{0.1},-\sqrt{0.1}$

2)Решим заменой => x^2-10=t

$t^2-3t-4=0\\D=9+4*4=9+16=25=5^2\\t_{1}=\frac{3+5}{2}=4\\t_{2}= \frac{3-5}{2}=-1$

a)

$x^2-10=4\\x^2=14\\x_{1}=\sqrt{14}\\x_{2}=-\sqrt{14}$

б)

$x^2-10=-1\\x^2=9\\x_{3}=3\\x_{4}=-3$

Ответ: 3 ; -3 ; $\sqrt{14},-\sqrt{14}$

3) Замена x^2=y

$y^2-10y+9=0\\D=100-4*9=100-36=64=8^2\\y_{1}=\frac{10+8}{2}=9\\y_{2}= \frac{10-8}{2}=1$

x= { ±1;±3}

Ответ: 1 ; -1 ; 3 ; -3

4)Группировка

$14x^2(x-2)-(x-2)=0\\(x-2)(14x^2-1)=0\\x=2\\14x^2=1\\x^2=\frac{1}{14}$

x=±$\sqrt{\frac{1}{14} }$

Ответ: 2 ; $\sqrt{\frac{1}{14} }$ ; -$\sqrt{\frac{1}{14} }$

5) замена x^2-7=t

$t^2-4t-5=0\\D=16+4*5=36=6^2\\t_{1}=\frac{4+6}{2}=5\\t_{2}=\frac{4-6}{2}= -1\\x^2-7=t$

а)

$x^2-7=5\\x^2=12\\x_{1}=\sqrt{12}\\x_{2}=-\sqrt{12}$

б)

$x^2-7=-1\\x^2=6\\x_{3}=\sqrt{6}\\x_{4}=-\sqrt{6}$

Ответ: $\sqrt{12}; -\sqrt{12} ; \sqrt{6} ; -\sqrt{6}$

6) x^2=y

$y^2-8y+7=0\\D=64-4*7=64-28=36=6^2\\y_{1}=\frac{8+6}{2}=7\\y_{2}=\frac{8-6}{2}=1$

x={±$\sqrt{7}$ ; ±1}

Ответ : 1 ; -1 ; $\sqrt{7}; -\sqrt{7}$

Answer 2

$1) \ 10x^3-20x^2-x+2=0\\10x^2(x-2)-(x-2)=0\\(x-2)(10x^2-1)=0\\\left \{ {{x-2=0} \atop {10x^2-1=0}} \right. =>\left \{ {{x_1=2} \atop {10x^2=1}} \right. =>x^2=\frac{1}{10}\\\\\\x_1=2\\x_{2,3}= \pm\sqrt{\frac{1}{10} }$

$2) \ (x^2-10)^2-3(x^2-10)-4=0\\x^2-10=t\\t^2-3t-4=0\\D=9+16=25=5^2\\t_1=(3-5)/2=-1\\t_2=(3+5)/2=4\\\\\left \{ {{x^2-10=-1} \atop {x^2-10=4}} \right. =>\left \{ {{x^2=9} \atop {x^2=14}} \right. =>\left \{ {{x_{1,2}=\pm3} \atop {x_{3,4}=\pm\sqrt{14} }} \right.$

$3) \ x^4-10x^2+9=0\\x^2=t\\t^2-10t+9=0\\D=100-36=64=8^2\\t_1=(10-8)/2=1\\t_2=(10+8)/2=9\\\\\left \{ {{x^2=1} \atop {x^2=9}} \right. =>\left \{ {{x_{1,2}=\pm1} \atop {x_{3,4}=\pm3}} \right.$

$5) \ (x^2-7)^2-4(x^2-7)-5=0\\x^2-7=t\\t^2-4t-5=0\\D=16+20=36=6^2\\t_1=(4-6)/2=-1\\t_2=(4+6)/2=5\\\\\left \{ {{x^2-7=-1} \atop {x^2-7=5}} \right. =>\left \{ {{x^2=6} \atop {x^2=12}} \right. =>\left \{ {{x_{1,2}=\pm\sqrt{6} } \atop {x_{3,4}=\pm2\sqrt{3} }} \right.$

$4) \ 14x^3-28x^2-x+2=0\\14x^2(x-2)-(x-2)=0\\(x-2)(14x^2-1)=0\\\\\left \{ {{x-2=0} \atop {14x^2-1=0}} \right. =>\left \{ {{x_1=2} \atop {x^2=\frac{1}{14} }} \right. =>\left \{ {{x_1=2} \atop {x_{2,3}=\pm\sqrt{\frac{1}{14} } }} \right.$

$6) \ x^4-8x^2+7=0\\x^2=t\\t^2-8t+7=0\\D=64-28=36=6^2\\t_1=(8-6)/2=1\\t_2=(8+6)/2=7\\\\\left \{ {{x^2=1} \atop {x^2=7}} \right. =>\left \{ {{x_{1,2}=\pm1} \atop {x_{3,4}=\pm\sqrt{7} }} \right.$