Question

Giải pt:

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

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

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

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

e) \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)

f) \(4x\sqrt{x+7}+3x\sqrt{7x-3}=6x^2+2\sqrt{7x^2+46x-21}\)

Answer 1

a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

Answer 2

b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)

Answer 3

c)ĐKXĐ:x=0 hoặc \(x\ge1;x\le-2\)

Nếu x=0=> VT=VP=0=> x=0 là 1 n_o

Nếu \(x\ge1.\)pt<=>\(\sqrt{x-1}+\sqrt{x+2}=2\sqrt{x}\Leftrightarrow x-1+x+2+2\sqrt{x^2+x-2}=4x\Leftrightarrow2x-1=2\sqrt{x^2+x-2}\Leftrightarrow4x^2-4x+1=4\left(x^2+x-2\right)\left(Dox\ge1\right)\)\(\Leftrightarrow8x=9\)=>....

Nếu \(x\le-2.\)Chia cả 2 vế của pt cho \(\sqrt{-x}\).Giải tương tự x>=1

Answer 4

d)ĐKXĐ: \(x\ge-1\)

pt<=>\(\sqrt{\dfrac{x^3+1}{x+3}}-\sqrt{x+3}=\sqrt{x^2-x+1}-\sqrt{x+1}\).

Bình phương 2 vế ta có: \(\dfrac{x^3+1}{x+3}+x+3-2\sqrt{x^3+1}=x^2-x+1+x+1-2\sqrt{x^3+1}\Leftrightarrow\left(x^2-x-1\right)\left(x+3\right)=x^3+1\Leftrightarrow x^3-x^2-x+3x^2-3x-3=x^3+1\Leftrightarrow2x^2-4x-4=0\)Tự giải chỗ nào bình phương thì thử lại

Answer 5

e,f ko cần làm cx đc