Question

câu hỏi ở phần trả lời

tại mk lười chép quá mong các bn thông cảm và giải hộ mk vs nhé

cảm ơn nhiều

Answer 1

CM CĐTS:

1)(\(\dfrac{\sqrt{14}-\sqrt{7}}{2\sqrt{2}-2}+\dfrac{\sqrt{15}-\sqrt{5}}{2\sqrt{3}-2}\)):\(\dfrac{1}{\sqrt{7}-\sqrt{5}}=1\)

<=>[\(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\)]:\(\dfrac{1}{\sqrt{7}-\sqrt{5}}=1\)

<=>\(\left(\dfrac{\sqrt{7}}{2}+\dfrac{\sqrt{5}}{2}\right).\left(\sqrt{7}-\sqrt{5}\right)=1\)

<=>\(\dfrac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{2}=1\)

<=>\(7-5=2\)

<=>2=2(luôn đúng)

2)\(\dfrac{4}{3+\sqrt{5}}+\dfrac{8}{\sqrt{5}-1}-\sqrt{\left(2-\sqrt{5}\right)^2}=7\)

<=>\(\dfrac{8}{6+2\sqrt{5}}+\dfrac{8}{\sqrt{5}-1}+\left(2-\sqrt{5}\right)=7\)

<=>\(\dfrac{8}{\left(\sqrt{5}+1\right)^2}+\dfrac{8}{\left(\sqrt{5}-1\right)}+2-\sqrt{5}=7\)

<=>\(\dfrac{8\left(\sqrt{5}-1\right)+8\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)}+2-\sqrt{5}=7\)

<=>\(\dfrac{40+24\sqrt{5}}{4\left(\sqrt{5}+1\right)}+2-\sqrt{5}=7\)

<=>\(\dfrac{4\left(10+6\sqrt{5}\right)}{4\left(\sqrt{5}+1\right)}-\sqrt{5}=5\)

<=>\(\dfrac{10+6\sqrt{5}}{\sqrt{5}+1}-\sqrt{5}=5\)

<=>\(\dfrac{10+6\sqrt{5}-5-\sqrt{5}}{\sqrt{5}+1}=5\)

<=>\(\dfrac{5+5\sqrt{5}}{\sqrt{5}+1}=5\)

<=>\(5=5\)(đúng)

l

Answer 2

1)

<=> A=\(\sqrt{3+2\sqrt{3}+1}-\sqrt{4+2.2.\sqrt{3}+3}\)

<=>\(A=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

<=>\(A=\left|\sqrt{3}+1\right|-\left|2+\sqrt{3}\right|\)

<=>\(A=\sqrt{3}+1-2-\sqrt{3}\)

<=>\(A=-1\)

B=\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

<=>\(B=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2.\sqrt{3}.\sqrt{2}+2}\)

<=>\(B=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

<=>\(B=\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\)

<=>\(B=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

<=>\(B=2\sqrt{2}\)

C=\(\sqrt{x+2-2\sqrt{x+1}}+\sqrt{x+2+2\sqrt{x+1}}\)

ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)

\(\Leftrightarrow C=\sqrt{\left(x+1\right)-2\sqrt{x+1}+1}+\sqrt{\left(x+1\right)+2\sqrt{x+1}+1}\)\(\Leftrightarrow C=\sqrt{\left(\sqrt{x+1}-1\right)^2}+\sqrt{\left(\sqrt{x+1}+1\right)^2}\)

\(\Leftrightarrow C=\left|\sqrt{x+1}-1\right|+\left|\sqrt{x+1}+1\right|\)

TH₁: \(-1\le x< 0\)

\(\Leftrightarrow C=-\sqrt{x+1}+1+\sqrt{x+1}+1\)

\(\Leftrightarrow C=2\)

TH₂:\(x\ge0\)

\(\Leftrightarrow C=\sqrt{x+1}-1+\sqrt{x+1}+1\)

\(\Leftrightarrow C=2\sqrt{x+1}\)

Vậy khi \(-1\le x< 0\)thì\(\Leftrightarrow C=2\)

KHi \(x\ge0\)thì \(\Leftrightarrow C=2\sqrt{x+1}\)

Answer 3