## A = – √2 + √(√ 2 – 1) ² A = 1/ x – 1 – 1/ x+1 + 4x + 2/x ² – 1

Question

A = – √2 + √(√ 2 – 1) ²
A = 1/ x – 1 – 1/ x+1 + 4x + 2/x ² – 1

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15 phút 2021-09-20T08:42:19+00:00 2 Answers 0 views 0

1. Giải thích các bước giải:

a)
A=-sqrt{2}+sqrt{(sqrt{2}-1)^2}
A=-sqrt{2}+sqrt{2}-1
A=(-sqrt{2}+sqrt{2})-1
A=0-1
A=-1

Vậy A=-1
b)
A=1/(x-1)-1/(x+1)+(4x+2)/(x^2-1)(x\ne+-1)
A=(1.(x+1))/((x-1).(x+1))-(1.(x-1))/((x+1).(x-1))
A=(x+1)/(x^2-1^2)-(x-1)/(x^2-1^2)+(4x+2)/(x^2-1)
A=(x+1)/(x^2-1)-(x-1)/(x^2-1)+(4x+2)/(x^2-1)
A=((x+1)-(x-1)+(4x+2))/(x^2-1)
A=(x+1-x+1+4x+2)/(x^2-1)
A=((x-x+4x)+(1+1+2))/(x^2-1)
A=(x(1-1+4)+4)/(x^2-1)
A=(x.4+4)/(x^2-1)
A=(4.(x+1))/(x^2-1)
A=(4.(x+1))/((x-1).(x+1))
A=4/(x-1)

Vậy A=4/(x-1)

2. Đáp án:$A=-1$

$A=\dfrac{4}{x-1}$

Giải thích các bước giải:

$A=-\sqrt{2}+\sqrt{(\sqrt{2}-1)^2}$

$A=-\sqrt{2}+\sqrt{2}-1$

$A=-1$

——————————————————————

$A=\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{4x+2}{x^2-1}$

$Đkxđ:x\neq \pm 1$

$A=\dfrac{x+1}{x^2-1}-\dfrac{x-1}{x^2-1}+\dfrac{4x+2}{x^2-1}$

$A=\dfrac{4x+4}{x^2-1}$
$A=\dfrac{4}{x-1}$