ĐK:\(x\ge3\)

PT \(\Leftrightarrow\frac{-6x}{\sqrt{x-3}+\sqrt{7x-3}}=\sqrt{5x-2}\)(nhân liên hợp)

Đến đây ta có VT < 0 với mọi \(x\ge3\) mà VP > 0. Vậy pt vô nghiệm.

đề sai r,,,,,,cái kia phải là x^2-x+1 chứ

nếu đúng như tôi thì bạn chỉ cần cho cái 2 vào trong căn rồi nhân liên hợp là ok

yes..thanks

\(\Leftrightarrow\dfrac{x+3+x-1+2\sqrt{\left(x+3\right)\left(x-1\right)}}{x+3-x+1}=\dfrac{13-x^2}{4}\)

\(\Leftrightarrow2x+2+2\sqrt{\left(x+3\right)\left(x-1\right)}=13-x^2\)

\(\Leftrightarrow\sqrt{4\left(x+3\right)\left(x-1\right)}=13-x^2-2x-2=-x^2-2x+11\)

=>\(x\simeq1,37\)

\(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)

\(\Leftrightarrow\sqrt[3]{x+3}-2-\left(\sqrt[3]{6-x}-1\right)=0\)

\(\Leftrightarrow\dfrac{x+3-8}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}-\dfrac{6-x-1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)

\(\Leftrightarrow\dfrac{x-5}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{x-5}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}\right)=0\)

Dễ thấy: \(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}>0\)

\(\Rightarrow x-5=0\Leftrightarrow x=5\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+3}=a\\\sqrt[3]{6-x}=b\end{matrix}\right.\)thì co hệ

\(\left\{{}\begin{matrix}a=1+b\left(1\right)\\a^3+b^3=9\left(2\right)\end{matrix}\right.\)

\(\Rightarrow\left(1+b\right)^3+b^3=9\)

\(\Leftrightarrow\left(b-1\right)\left(2b^2+5b+8\right)=0\)

Dễ thây \(2b^2+5b+8>0\)

\(\Rightarrow b=1\)

\(\Rightarrow\sqrt[3]{6-x}=1\)

\(\Leftrightarrow x=5\)

\(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)

\(\Leftrightarrow\sqrt[3]{x+3}-2-\left(\sqrt[3]{6-x}-1\right)=0\)

\(\Leftrightarrow\frac{x+3-8}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}-\frac{6-x-1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\frac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}\right)=0\)

Dễ thấy :

\(\frac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\frac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}>0\)

\(\Rightarrow x-5=0\Leftrightarrow x=5\)

Chúc bạn học tốt !!!

Áp dụng BĐT:\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Ta có: \(\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)

Dấu \(=\)xảy ra khi \(AB\ge0\)

dat \(\sqrt{x-1}\) = t

ta có: \(\sqrt{x+3+4t}\)+ \(\sqrt{x+8-6t}\)= 5

     x + 3 + 4t + x + 8 - 6t = 25

   2x - 2t = 14 ( chia cả 2 vế cho 2)

   x - t = 7

   t = x - 7

  thay t = \(\sqrt{x}-1\)vào ta được:

 x - 7 = \(\sqrt{x-1}\)

( x - 7 )² = x - 1

x² -14x + 49 = x - 1

x² - 15x + 50 = 0

​k biết đúng hay k

liên hợ thôi !

\(DK:x\ge0\)

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

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

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

\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)

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

\(\Leftrightarrow x=1\)

Vay nghiem cua PT la \(x=1\)