mình có sửa lại đề 1 chút!

đặt \(T=\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

đặt \(u=a^4;v=b^6\)(a,b>0) ta có

\(T=\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}=\frac{a^4-8a^2b^2+4b^2}{a^2-2b^2+2ab}+3b^2\)

vậy \(T=\frac{a^4-8a^2b^2+4b^4}{a^2-2b^2+2ab}+3b^2=\frac{a^4-5a^2b^2-2b^4+6ab^3}{a^2-2b^2+2ab}=a^2-2ab+b^2\)

từ đó suy ra \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}\)

vì \(u^3\ge v^2\)nên \(\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}=\sqrt[4]{u}\)

\(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

với u=1 ta có \(T=\sqrt{\frac{1-8\sqrt[6]{v^2}+4\sqrt[3]{v^2}}{1-2\sqrt[3]{v}+2\sqrt[6]{v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}\)

nếu \(1-2\sqrt[3]{v}+2\sqrt[6]{v}=0\)thì \(\sqrt[3]{v}=\frac{3+1}{2}>0\)

do \(v^2>1=u^3\), mâu thuẫn suy ra \(1-2\sqrt[3]{v}+2\sqrt[6]{v}\ne0\)

tóm lại với \(u^3\ge v^2\)và u,v\(\inℚ^+\)để \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)cần và đủ là u=1 và v<1, v\(\inℚ^+\)được lấy tùy ý

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=\left(\sqrt{3-\sqrt{5}}\right)^2+2\sqrt{\left(3-\sqrt{5}\right).\left(3+\sqrt{5}\right)}+\left(\sqrt{3+\sqrt{5}}\right)^2\)

\(=3-\sqrt{5}+2.\sqrt{3^2-5}+3+\sqrt{5}\)

\(=6+2.\sqrt{4}=6+2.2=6+4=10\)

hình như kết quả là 5 í!

Kết quả là 25

Bài làm:

Ta có: \(M=\sqrt{x^2+2x+5}=\sqrt{\left(x+1\right)^2+4}\)

Mà \(\left(x+1\right)^2+4\ge4\left(\forall x\right)\)

=> \(M\ge2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy \(M_{Min}=2\Leftrightarrow x=-1\)

\(M=\sqrt{x^2+2x+5}\)

\(\Leftrightarrow M=\sqrt{x^2+2x+1+4}\)

\(\Leftrightarrow M=\sqrt{\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Min M = 2 

\(\Leftrightarrow x=-1\)

Bài làm:

Ta có: \(\sqrt{3-2x}< 5\)

\(\Leftrightarrow\left|3-2x\right|< 25\)

\(\Leftrightarrow-5< 3-2x< 5\)

\(\Leftrightarrow3-\left(-5\right)>3-\left(3-2x\right)>3-5\)

\(\Leftrightarrow8>2x>-2\)

\(\Rightarrow-1< x< 4\)

Đặt \(x^2=t\left(t\ge0\right)\)

\(\Leftrightarrow t^2-16t+32=0\)

\(\Delta=\left(-16\right)^2-4.32=256-128=128>0\)

\(t_1=\frac{16-\sqrt{128}}{2}=8-4\sqrt{2};t_2=\frac{16+\sqrt{128}}{2}=8+4\sqrt{2}\)

Theo bài ra ta có : 

\(x_0=\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\)

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

tịt lun, cái pt căn này chill quá 

 ๖²⁴ʱ๖ۣۜTɦủү❄吻༉ Mơn Bạn nha .

P/s : làm nháp thử mn sửa giúp nha ( thực ra em cũng chả hiểu cái gì cả T_T )

Ta có :

\(\left(x_0\right)^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2-\sqrt{3}\right)}\)

\(\Rightarrow\left(\frac{8-\left(x_0\right)^2}{2}\right)^2=2+\sqrt{3}+3\left(2-\sqrt{3}\right)+2\sqrt{3\left(4-3\right)}=8\)

\(\Rightarrow64-16\left(x_0\right)^2+\left(x_0\right)^4=32\)

\(\Rightarrow\left(x_0\right)^4-16\left(x_0\right)^2+32=0\left(đpcm\right)\)

Đặt: \(A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\)

=> \(A^2=\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

=> \(A^2=2\sqrt{5}+2\sqrt{5-4}\)

=> \(A^2=2\sqrt{5}+2\)

=> \(A^2=2\left(\sqrt{5}+1\right)\)

=> \(A=\sqrt{2\left(\sqrt{5}+1\right)}\)

=> \(\frac{A}{\sqrt{\sqrt{5}+1}}=\frac{\sqrt{2\left(\sqrt{5}+1\right)}}{\sqrt{\sqrt{5}+1}}=\sqrt{2}\)

Đặt: \(B=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

=> \(VT=\frac{A}{\sqrt{\sqrt{5}+1}}-B=\sqrt{2}-\left(\sqrt{2}-1\right)=\sqrt{2}-\sqrt{2}+1=1\)

VẬY KẾT QUẢ CỦA PHÉP TÍNH = 1.

gt <=> \(\sqrt{x-1}-\sqrt{y-1}+\left(x-y\right)\left(x+y\right)=0\)

<=> \(\frac{\left(\sqrt{x-1}-\sqrt{y-1}\right)\left(\sqrt{x-1}+\sqrt{y-1}\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\) 

<=> \(\frac{\left(x-1\right)-\left(y-1\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\)

<=> \(\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\)

<=> \(\left(x-y\right)\left(x+y+\frac{1}{\sqrt{x-1}+\sqrt{y-1}}\right)=0\)      (1)

Mà theo ĐKXĐ thì: \(x;y\ge1\)

=> \(x+y\ge2>0\)

Mà \(\frac{1}{\sqrt{x-1}+\sqrt{y-1}}>0\)

=> \(x+y+\frac{1}{\sqrt{x-1}+\sqrt{y-1}}>0\)       (2)

Từ (1) và (2) thì: 

=> \(x=y\)

VẬY TA CÓ ĐPCM.