Đặt \(130307=a;\text{ }140307=b\)

Pt trở thành \(\sqrt{a+b\sqrt{x+1}}=1+\sqrt{a-b\sqrt{x+1}}\)

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

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

\(\Leftrightarrow2a-1=2\sqrt{a^2-b^2\left(x+1\right)}\)

\(\Leftrightarrow\left(2a-1\right)^2=4\left[a^2-b^2\left(x+1\right)\right]\)

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

\(\Leftrightarrow x=\frac{4a^2-\left(2a-1\right)^2}{4b^2}-1\)

ĐKXĐ : \(\left\{{}\begin{matrix}-130307\le140307\sqrt{1+y}\\130307\ge140307\sqrt{1+y}\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{1+y}\le\dfrac{130307}{140307}\) và \(y\ge-1\)

\(PT\Leftrightarrow140307\sqrt{1+y}=-140307\sqrt{1+y}\)

\(\Leftrightarrow\)\(\sqrt{1+y}=0\)

\(\Leftrightarrow y=-1\) ( TM )

Vậy ...
 

giúp em với mọi người ơi em đang cần gấp lắm ạ TT

sai đề

\(DK:x\ge0\)

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

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

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

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

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

\(\Leftrightarrow x=1\)

Vay nghiem cua PT la \(x=1\)

\(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0;ĐK:x\ge4\)

\(\Leftrightarrow\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}-\sqrt{x+4}\)

\(\Leftrightarrow2x+9+2\sqrt{x^2+9x}=2x-5+2\sqrt{x^2-5x+4}\)

\(\leftrightarrow14+2\sqrt{x^2+9x}=2\sqrt{x^2-5x+4}\leftrightarrow7+\sqrt{x^2+9x}=\sqrt{x^2-5x+4}\)

\(\leftrightarrow49+14\sqrt{x^2+9x}+x^2+9x=x^2-5x+4\)

\(\leftrightarrow14\sqrt{x^2+9x}=-14x-45\)

\(\leftrightarrow\hept{\begin{cases}196.x^2+9x=196x^2+1260x+2025\\-14x-45\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}504x=2025\\x\le\frac{-45}{14}\end{cases}\leftrightarrow x=\frac{225}{56}}\) loại

-> PT vô nghiệm

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

OoO Ledegill2 OoO. Ban co the giai thich ro hon giup minh duoc khong. hi

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

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

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

=>\(x\simeq1,37\)