a) \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)=3

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

<=> x+1+x-2=3

<=> 2x = 3+2-1

<=>2x=4

<=> x=2

ĐK: \(x^3+4x^2+5x+6\ge0\)

Ta có: \(x^3+4x^2+5x+6=\left(x+3\right)\left(x^2+x+2\right);x^2+2x+5=\left(x+3\right)+\left(x^2+x+2\right)\)

Đặt \(\hept{\begin{cases}\sqrt{x+3}=u\\\sqrt{x^2+x+2}=v\end{cases}}\)

Vậy nên ta có phương trình: \(\)\(u^2+v^2=\frac{5}{2}uv\)

\(\Leftrightarrow2u^2-5uv+2v^2=0\Leftrightarrow\orbr{\begin{cases}u=2v\\u=\frac{1}{2}v\end{cases}}\)

Với u = 2v ta có: \(\sqrt{x+3}=2\sqrt{x^2+x+2}\Leftrightarrow x+3=4x^2+4x+8\)

\(\Leftrightarrow4x^2+3x+5=0\)   (Vô nghiệm)

Với \(u=\frac{1}{2}v\) ta có: \(2\sqrt{x+3}=\sqrt{x^2+x+2}\Leftrightarrow4x+12=x^2+x+2\)

\(\Leftrightarrow x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\left(tmđk\right)\)

Vậy phương trình có nghiệm \(x\in\left\{5;-2\right\}\)

Chu vi \(\Delta ABC\)là :

105 + 4 + 45 = 154

 Đ/s:.................

Làm:

Chu vi \(\Delta ABC\)là:

\(105+4+45=154\)

Đ/s:.........

~ học tốt ~

a) Đk: x \(\ge\) 5

\(\sqrt{x-5}-\frac{x-14}{3x+\sqrt{x-5}}=3\)

\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\frac{x-14}{3\sqrt{x-3}}\left(3+\sqrt{x-5}\right)=3\left(3+\sqrt{x-5}\right)\)

\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\left(x-14\right)=3\left(3+\sqrt{x-5}\right)\)

\(3\sqrt{x-5}+9-\left(3\sqrt{x-5}+9\right)=9+3\sqrt{x-5}-\left(3\sqrt{x-5}+9\right)\)

=> Luôn đúng với x \(\ge\) 5

chúc bạn học tốt 

Ai còn onl ko kb vs mk bùn quá!!!

Đặt:   \(A=\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}\)\(>\)\(0\)

=>   \(A^2=\frac{7+\sqrt{5}+2.\sqrt{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}+7-\sqrt{5}}{7+2\sqrt{11}}\)

            \(=\frac{14+4\sqrt{11}}{7+2\sqrt{11}}\)

             \(=\frac{2\left(7+2\sqrt{11}\right)}{7+2\sqrt{11}}=2\)

=>  \(A=\sqrt{2}\)

\(D=\sqrt{2}-\sqrt{3-2\sqrt{2}}\)

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

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

giúp mình với

\(A=\frac{\sqrt{2}}{\sqrt{2}+\sqrt{2+\sqrt{2}}}+\frac{\sqrt{2}}{\sqrt{2}+\sqrt{2-\sqrt{2}}}\)

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

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

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

\(=\sqrt{2+\sqrt{2}}-\sqrt{2-\sqrt{2}}\)

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

            \(=4-2.\sqrt{2}\)

=>  \(A=\sqrt{4-2\sqrt{2}}\)

P/S: mk lm đc có z thôi, bn tham khảo, sai đâu m.n bỏ qua