ĐK : \(x\ge1\)

\(A=\sqrt{x+2\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}\)

\(=\sqrt{x-1+2\sqrt{x-1}}-\sqrt{x-1+6\sqrt{x-1}+9}\)

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

\(=\left|\sqrt{x-1}-1\right|-\left|\sqrt{x-1}+3\right|\)

\(=\hept{\begin{cases}1-\sqrt{x-1}-\sqrt{x-1}-3;1\le x\le2\\\sqrt{x-1}-1-\sqrt{x-1}-3;x>2\end{cases}}\)

\(=\hept{\begin{cases}-2-2\sqrt{x-1};1\le x\le2\\-4;x>2\end{cases}}\)

bạn viết linh tinh j vậy ạ???

bạn bấm máy giải phương trình bậc 2 

hoặc đưa về phương trình \(A^2=B^2\)như sau:\(x^2+4x-2=0\)

\(x^2+2.x.2+2^2-6=0\)

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

\(\left|x+2\right|=\sqrt{6}\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{6}\\x+2=-\sqrt{6}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-2\\x=-\sqrt{6}+2\end{cases}}\)

\(ĐKXĐ:x;y\ge2\)

Trừ 2 vế của hệ cho nhau ta được

\(\left(\sqrt{x+1}-\sqrt{y+1}\right)+\left(\sqrt{y-2}-\sqrt{x-2}\right)=0\)

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

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

\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{x+1}+\sqrt{y+1}}-\frac{1}{\sqrt{x-2}+\sqrt{y-2}}\right)=0\)(1)

Vì \(\sqrt{x+1}+\sqrt{y+1}>\sqrt{x-2}+\sqrt{y-2}\)

\(\Rightarrow\frac{1}{\sqrt{x+1}+\sqrt{y+1}}< \frac{1}{\sqrt{x-2}+\sqrt{y-2}}\)

\(\Rightarrow\frac{1}{\sqrt{x+1}+\sqrt{y+1}}-\frac{1}{\sqrt{x-2}+\sqrt{y-2}}< 0\)(2)

Từ (1) và (2) => x - y = 0

                    <=> x = y

Thay vào 1 trong 2 pt ban đầu có

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

\(\Leftrightarrow x+1+2\sqrt{\left(x+1\right)\left(x-2\right)}+x-2=9\)

\(\Leftrightarrow\sqrt{x^2-x-2}=5-x\)

\(\Leftrightarrow\hept{\begin{cases}x\le5\\x^2-x-2=25-10x+x^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le5\\9x=27\end{cases}}\)

\(\Leftrightarrow x=3\left(tmĐKXĐ\right)\)

Vậy pt có nghiệm duy nhất x = 3

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2.\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}\)

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

=> \(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k\left(k+1\right)+1}{k\left(k+1\right)}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)

=> \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}\)

\(=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{k}-\frac{1}{k+1}\)

\(=k+1-\frac{1}{k+1}\)

=> \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{2017^2-1}{2017}\)

<=> \(k+1-\frac{1}{k+1}=2017-\frac{1}{2017}\)

\(\Leftrightarrow\left(k+1-2017\right)-\left(\frac{1}{k+1}-\frac{1}{2017}\right)=0\)

\(\Leftrightarrow\left(k-2016\right)\left(1+\frac{1}{2017.\left(k+1\right)}\right)=0\)

<=> k=2016

Câu hỏi của phạm trung hiếu - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo nhé!

\(a,\frac{6-3\sqrt{6}}{6}=\frac{3\left(2-\sqrt{6}\right)}{6}=\frac{2-\sqrt{6}}{2}\)

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

\(b,ĐKXĐ:x>0\)

\(D=2011\sqrt{x}-2+\frac{1}{\sqrt{x}}\)\(=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Áp dụng bđt Cauchy cho 2 số dương \(2011\sqrt{x}\)và\(\frac{1}{\sqrt{x}}\)ta được:

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}.\frac{1}{\sqrt{x}}}\)

\(\Leftrightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{2011}-2\)

\(\Leftrightarrow D\ge2\sqrt{2011}-2\)

Dấu "=" xảy ra \(\Leftrightarrow2011\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2011}\left(TMĐK\right)\)

^{\(a,\sqrt{2x-1}=2\)}

\(\Rightarrow2x-1=4\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=\frac{5}{2}\)

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

\(\Rightarrow2x-1=\left(x+1\right)^2\)

\(\Rightarrow2x-1=x^2+2x+1\)

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

\(\Rightarrow x^2=-2VN\)