Đáp án: $\text{Ảnh 1:}$`d)` `(1-sqrta)/(2sqrta-1)`;
$\text{Ảnh 2: 1) b, 1; d, 2015 }$
Giải thích các bước giải:
$\text{Ảnh 1:}$ `d)` `(3sqrta-2a-1)/(4a-4sqrta+1)=(2sqrta+sqrta-2a-1)/((2sqrta)^2-2.2sqrta.1+1^2)=((2sqrta-1)+(sqrta-2a))/(2sqrta-1)^2`
`=((2sqrta-1)-sqrta(-1+2sqrta))/(2sqrta-1)^2=((2sqrta-1)-sqrta(2sqrta-1))/((2sqrta-1)(2sqrta-1))`
`=((2sqrta-1)(1-sqrta))/((2sqrta-1)(2sqrta-1))=(1-sqrta)/(2sqrta-1)`
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$\text{Ảnh 2:}$ `1)` `b)` `(sqrt(3-sqrt5).(3+sqrt5))/(sqrt(10)+sqrt2)=(sqrt(3-sqrt5).sqrt(3+sqrt5).sqrt(3+sqrt5))/(sqrt(10)+sqrt2)`
`=(sqrt((3-sqrt5)(3+sqrt5)).sqrt(3+sqrt5))/(sqrt(10)+sqrt2)=(sqrt(9-5).sqrt(3+sqrt5))/(sqrt(10)+sqrt2)`
`=(2.sqrt(3+sqrt5))/(sqrt(10)+sqrt2)=(sqrt(4.3+4.sqrt5))/(sqrt(10)+sqrt2)=(sqrt(12+4.sqrt5))/(sqrt(10)+sqrt2)=(sqrt(10+2.sqrt20+2))/(sqrt(10)+sqrt2)`
`=(sqrt(sqrt(10)^2+2.sqrt(10.2)+sqrt2^2))/(sqrt(10)+sqrt2)=(sqrt(sqrt(10)+sqrt2)^2)/(sqrt(10)+sqrt2)=(sqrt(10)+sqrt2)/(sqrt(10)+sqrt2)=1`
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`1) d)` `sqrt((1-sqrt(2016))^2).sqrt(2017+2sqrt2016)`
`=|1-sqrt(2016)|.sqrt(2016+2sqrt2016+1)`
`=|1-sqrt(2016)|.sqrt(sqrt2016^2+2sqrt2016.1+sqrt1^2)`
`=|1-sqrt(2016)|.sqrt((sqrt2016+1)^2)=|1-sqrt(2016)|.|sqrt2016+1|`
`=(sqrt(2016)-1).(sqrt2016+1)=sqrt(2016)^2-1^2=2016-1=2015`
