\(b\ge9,a\ge\sqrt{b}\)

a, 

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1 tháng 6 2021

a) 

\(\left(\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}\right)^2\)

\(=a+\sqrt{b}\ne2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}+a-\sqrt{b}\)

\(=2a\ne2\sqrt{a^2-b}=2\left(a\ne\sqrt{a^2}-b\right)\)

\(\Rightarrow\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\ne\sqrt{a^2}-b\right)}\)

\(\Rightarrowđpcm\)

1 tháng 6 2021

b)

\(\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}\ne}\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2\)

\(=\frac{a+\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a+\sqrt{a^2-b}}{2}.\frac{a-\sqrt{a^2-b}}{2}}+\frac{a-\sqrt{a^2-b}}{2}\)

\(=\frac{a}{2}+\frac{\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a^2-a^2+b}{2.2}}+\frac{a}{2}-\frac{\sqrt{a^2-b}}{2}\)

\(=a\ne2\frac{\sqrt{b}}{2}=a\ne\sqrt{b}\)

\(\Rightarrow\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\ne\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{a\ne\sqrt{b}}\)

\(\Rightarrowđpcm\)

21 tháng 9 2018

Với b\(\ge\)0, a\(\ge\)\(\sqrt{b}\) ta bình phương 2 vế lên có:

\(\sqrt{a\pm \sqrt{b}}^2\)=\((\sqrt{\dfrac{\sqrt{a+\sqrt{a^2-b}}}{2}}\)\pm \(\sqrt{\dfrac{\sqrt{a-\sqrt{a^2-b}}}{2}})^2\)

21 tháng 9 2018

Xét vế trái ta có:

\(\sqrt{(a\pm \sqrt{b})^2}\)=\(a\pm \sqrt{b})

26 tháng 9 2020

hộ mình câu c ạ :(((

AH
Akai Haruma
Giáo viên
13 tháng 9 2018

Lời giải:

Sửa đề: \(\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)


Xét

\(B=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)

\(B^2=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{a+\sqrt{a^2-b}}{2}.\frac{a-\sqrt{a^2-b}}{2}}\)

\(=a+2\sqrt{\frac{a^2-(a^2-b)}{4}}=a+\sqrt{b}\)

\(\Rightarrow B=\sqrt{a+\sqrt{b}}\) (do B không âm.)

Hoàn toàn tt, \(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{a-\sqrt{b}}\)

24 tháng 9 2020

1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)

2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)

\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)

4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)