Question

Giải phương trình :

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

b, \(x-\sqrt{2x+3}=0\)

c, \(\sqrt{x-2}-3\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

d, \(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)

e, \(2\sqrt{x+3}=9x^2-x-4\)

f, \(\sqrt{x+1}-\sqrt{x-7}=\sqrt{12-x}\)

g, \(\sqrt{2x+5}-\sqrt{3x-5}=2\)

h, \(\sqrt{x}-\sqrt{x-1}-\sqrt{x-4}+\sqrt{x+9}=0\)

i, \(x^2+2x-\sqrt{x^2+2x+1}-5=0\)

k, \(\sqrt{x+8-6\sqrt{x+1}}=4\)

l, \(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)

Làm được phần nào thì giúp mình nha đang cần gấp !!!

Answer 1

I) xd mọi x

\(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)

\(\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-5\right)^2}=9=>\left|x-4\right|+\left|x-5\right|=9\)

\(\left[{}\begin{matrix}x< 4\Rightarrow4-x+5-x=>x=0\left(n\right)\\4\le x< 5\Rightarrow x-4+5-x=9\left(vn\right)\\x\ge5\Rightarrow x-4+x-5=9\Rightarrow x=9\left(n\right)\\\end{matrix}\right.\)

kết luận

\(\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)