Ta có: \(\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2}}.\sqrt{4+2\sqrt{3}}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)

=> \(A=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}}=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}}=\frac{\sqrt{3}+1}{2}\)

a) \(\left(xy+1\right)^2=25\)

\(\Leftrightarrow\orbr{\begin{cases}xy+1=5\\xy+1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}xy=4\\xy=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{y}\\x=-\frac{6}{y}\end{cases}}\)

+ Nếu: \(x=\frac{4}{y}\Leftrightarrow\left(\frac{4}{y}+y\right)^2=49\)

\(\Leftrightarrow y^2+8+\frac{16}{y^2}=49\)

\(\Leftrightarrow\frac{y^4+16}{y^2}=41\)

\(\Leftrightarrow y^4-41y^2+16=0\) => y vô tỉ (loại)

+ Nếu: \(x=-\frac{6}{y}\Rightarrow\left(y-\frac{6}{y}\right)^2=49\)

\(\Leftrightarrow y^2+\frac{36}{y^2}=49+12\)

\(\Leftrightarrow y^4-61y^2+36=0\) => y vô tỉ (loại)

=> hpt vô nghiệm

b) tương tự

a) 

= \(\sqrt{18-6\sqrt{6}+3}\)        

= \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)       

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

= \(|3\sqrt{2}-\sqrt{3}|\)   

= \(3\sqrt{2}-\sqrt{3}\)   

b) 

= \(\sqrt{\frac{7}{2}-\sqrt{7}+\frac{1}{2}}\)   

= \(\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)    

= \(\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)     

= \(|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}|\) 

= \(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\)        

c) 

= \(\sqrt{3+2\sqrt{3}+1}\)  

= \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}\)    

= \(\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\) 

d) 

Đặt t = \(\sqrt{x-1}\left(ĐK:t\ge0\right)\)   

= \(\sqrt{t^2+1-2t}\)       

= \(\sqrt{\left(t+1\right)^2}\)   

\(=t+1\)      

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

\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2\sqrt{9}\sqrt{6}+3}=\sqrt{\left(\sqrt{18}\right)^2-2\sqrt{18}\sqrt{3}+\left(\sqrt{3}\right)^2}\)

                                \(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}=\sqrt{3}+3\sqrt{2}\)

\(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)

                           \(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{2}}{2}\)

\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Với \(x\ge1\)thì \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)

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

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

T đã tốn mấy phút cuộc đời viết lời giải cho bạn r, tiếc j mấy giây mà bấm k cho t ik =))

x; y nguyên

pt <=> \(29x^2=2+28y^2⋮2\) mà 29 không chia hết cho 2 => x² chia hết cho 2 => x chia hết cho 2

=> Tồn tại số nguyên k sao cho: x = 2k 

=> \(29.4k^2-28.y^2=2\)

<=> \(1=29.2k^2-14y^2\)chia hết cho 2 

=> Vô lí

=> pt ban đầu vô nghiệm 

giải được chết liền

what the hell

??????????????

a) Ta có: \(\sqrt{3x-2}=x+1\) ( ĐK: \(x\ge\frac{2}{3}\)) 

         \(\Leftrightarrow\left(\sqrt{3x-2}\right)^2=\left(x+1\right)^2\)

         \(\Leftrightarrow3x-2=x^2+2x+1\)

         \(\Leftrightarrow x^2-x+3=0\)

         \(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\frac{11}{4}=0\)

         \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)

    Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\forall x\)

          mà \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)

   \(\Rightarrow\)\(S=\varnothing\)

b) Ta có: \(\sqrt{x^2-2x+1}=x-1\)  ( ĐK: \(x\inℝ\))

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

        \(\Leftrightarrow\left|x-1\right|=x-1\)

   + Với \(x< 1\)\(\Rightarrow\)\(\left|x-1\right|=1-x\)

      Ta có: \(1-x=x-1\)

          \(\Leftrightarrow-2x=-2\)

          \(\Leftrightarrow x=1\left(L\right)\)

    + Với \(x\ge1\)\(\Rightarrow\)\(\left|x-1\right|=x-1\)

       Ta có: \(x-1=x-1\)

          \(\Leftrightarrow0x=0\)

          \(\Rightarrow\)\(x\inℝ\)

   Vậy \(S=ℝ\)

c) Ta có: \(\sqrt{2x+1}=x-2\) ( ĐK: \(x\ge-\frac{1}{2}\))

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

        \(\Leftrightarrow2x+1=x^2-4x+4\)

        \(\Leftrightarrow x^2-6x+3=0\)

        \(\Leftrightarrow\left(x^2-6x+9\right)-6=0\)

        \(\Leftrightarrow\left(x-3\right)^2=6\)

        \(\Leftrightarrow x-3=\pm\sqrt{6}\)

   + \(x-3=\sqrt{6}\)\(\Leftrightarrow\)\(x=3+\sqrt{6}\approx5,45\)\(\left(TM\right)\)

   + \(x-3=-\sqrt{6}\)\(\Leftrightarrow\)\(x=3-\sqrt{6}\approx0,55\)\(\left(TM\right)\)

   Vậy \(S=\left\{5,45;0,55\right\}\)

d) Ta có: \(\sqrt{x^2-3}=x^2-3\) ( ĐK: \(x\ge\pm\sqrt{3}\))

        \(\Leftrightarrow\sqrt{x^2-3}-\left(\sqrt{x^2-3}\right)^2=0\)

        \(\Leftrightarrow\sqrt{x^2-3}.\left(1-\sqrt{x^2-3}\right)=0\)

   + \(\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(x^2-3=0\)\(\Leftrightarrow\)\(x^2=3\)\(\Leftrightarrow\)\(x=\pm\sqrt{3}\)\(\left(TM\right)\)

   + \(1-\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(\sqrt{x^2-3}=1\)

                                                 \(\Leftrightarrow\)\(x^2-3=1\)

                                                 \(\Leftrightarrow\)\(x^2=4\)

                                                 \(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\left(TM\right)\\x=-2\left(L\right)\end{cases}}\)

   Vậy \(S=\left\{-\sqrt{3};\sqrt{3};2\right\}\)

e) Ta có: \(\sqrt{x^2-6x+9}=6-x\) ( ĐK: \(x\inℝ\))

        \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)

        \(\Leftrightarrow\left|x-3\right|=6-x\)

   + Với \(x< 3\)\(\Leftrightarrow\)\(\left|x-3\right|=3-x\)

       Ta có: \(3-x=6-x\)

           \(\Leftrightarrow0x=3\)( vô nghiệm )

   + Với \(x\ge3\)\(\Leftrightarrow\)\(\left|x-3\right|=x-3\)

       Ta có: \(x-3=6-x\)

          \(\Leftrightarrow2x=9\)

          \(\Leftrightarrow x=\frac{9}{2}\)\(\left(TM\right)\)

   Vậy \(S=\left\{\frac{9}{2}\right\}\)

g) Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\) ( ĐK: \(x\inℝ\))

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

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

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

   + Với \(x< 2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1< 0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)

      Ta có: \(1-\sqrt{x-1}=\sqrt{x-1}-1\)

           \(\Leftrightarrow-2\sqrt{x-1}=-2\)

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

           \(\Leftrightarrow x-1=1\)

           \(\Leftrightarrow x=2\)\(\left(L\right)\)

   + Với \(x\ge2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1\ge0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)

      Ta có: \(\sqrt{x-1}-1=\sqrt{x-1}-1\)

          \(\Leftrightarrow0x=0\)( vô số nghiệm )

   Vậy \(S=ℝ\)

a) Ta có: \(F=\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge\sqrt{1}=1\left(\forall x\right)\)

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

Vậy Min(F) = 1 khi x=2

b) \(D=\sqrt{2x^2-4x+10}=\sqrt{2\left(x-1\right)^2+8}\ge\sqrt{8}=2\sqrt{2}\left(\forall x\right)\)

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

Vậy \(Min\left(D\right)=2\sqrt{2}\Leftrightarrow x=1\)

c) \(G=\sqrt{2x^2-6x+5}=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}}\ge\sqrt{\frac{1}{2}}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy \(Min\left(G\right)=\frac{\sqrt{2}}{2}\Leftrightarrow x=\frac{3}{2}\)

đề bài đúng không z? theo tôi đề là \(\sqrt{x+2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)?!

ĐKXĐ:...

Áp dụng BĐT AM-GM:

\(\left(\sqrt{x+2}+\sqrt{6-x}\right)^2\le2\left(x+2+6-x\right)=16\)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{6-x}\le4\)

Lại có \(x^2-8x+24=\left(x-4\right)^2+8\ge8\forall x\)

Vậy pt vô nghiệm.

Các bạn giúp mk nhanh vs aaaaaasắp đến hạn nộp rồi

Bài làm:

Ta có: \(P=\frac{4}{a}+\frac{4}{b}+3a+3b-2\)

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

Áp dụng bất đẳng thức Cauchy ta được:

\(P\ge2\sqrt{\frac{4}{a}.a}+2\sqrt{\frac{4}{b}.b}+2.4-2\)

\(=4+4+8-2=14\)

Dấu "=" xảy ra khi: \(a=b=2\)

Vậy Min(P) = 14 khi a=b=2

Bài làm:

đk: \(\hept{\begin{cases}x\ge0\\y\ge0\end{cases}}\)

Ta thấy: \(\hept{\begin{cases}\sqrt{x}\ge0\left(\forall x\right)\\\sqrt{y}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\sqrt{x}+\sqrt{y}\ge0\left(\forall x,y\right)\Rightarrow\frac{\sqrt{x}+\sqrt{y}}{4}\ge0\left(\forall x,y\right)\)

=> \(A\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\sqrt{x}=0\\\sqrt{y}=0\end{cases}}\Rightarrow x=y=0\)

Vậy Min(A) = 0 khi x=y=0