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

a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

\(x+\sqrt{5+\sqrt{x-1}}=6\)

Đk:\(x\ge1\)

\(pt\Leftrightarrow\sqrt{5+\sqrt{x-1}}=6-x\)

\(\Leftrightarrow5+\sqrt{x-1}=x^2-12x+36\)

\(\Leftrightarrow\sqrt{x-1}=x^2-12x+31\)

\(\Leftrightarrow x-1=x^4-24x^3+206x^2-744x+961\)

\(\Leftrightarrow-x^4+24x^3-206x^2+745x-962=0\)

\(\Leftrightarrow-\left(x^2-13x+37\right)\left(x^2-11x+26\right)=0\)

\(\Rightarrow x=-\frac{\sqrt{17}-11}{2}\) (thỏa)

mk ko bt 123

Đặt \(\hept{\begin{cases}\sqrt[3]{2-x}=b\\\sqrt{x-1}=a\end{cases}}\)

Ta có hệ \(\hept{\begin{cases}a^2+b^3=1\\a-b=5\end{cases}}\)

<=> \(\hept{\begin{cases}a=3\\b=-2\end{cases}}\)

<=> x = 10

\(\sqrt{x+1}=\sqrt{5}-3\)

\(x+1=5-2.\sqrt{5}.3+9\)

\(x+1=14-6.\sqrt{5}\)

\(x=13-6.\sqrt{5}\)

ĐK: \(x+1\ge0\Leftrightarrow x\ge-1.\)

Bình phương hai vế

\(x+1=5+9-2.3.\sqrt{5}\Rightarrow x=13-6.\sqrt{5}\)

1; Khi m=1 thì pt sẽ là \(\sqrt{x+1}=x+1\)

=>(x+1)^2=(x+1)

=>x(x+1)=0

=>x=0hoặc x=-1

2: \(\Leftrightarrow x+1=\left(x+m\right)^2\)

=>x^2+2mx+m^2-x-1=0

=>x^2+x(2m-1)+m^2-1=0

Δ=(2m-1)^2-4(m^2-1)

=4m^2-4m+1-4m^2+4

=-4m+5

Để pt có 2 nghiệm pb thì -4m+5>0

=>-4m>-5

=>m<5/4

Để pt có nghiệm kép thì 5-4m=0

=>m=5/4

Để pt vô nghiệm thì -4m+5<0

=>m>5/4

sai đề