Question

giải phương trình:

a, \(x+4\sqrt{7-x}=4\sqrt{x-1}+\sqrt{\left(7-x\right)\left(x-1\right)}+1\)

b, \(x^2+2x+4=3\sqrt{x^2+4x}\)

Answer 1

a,\(x+4\sqrt{7-x}\) \(-4\sqrt{x-1}-\sqrt{\left(7-x\right)\left(x-1\right)}-1=0\) (dk \(1\le x\le7\) )

\(\Leftrightarrow\left(\sqrt{x-1}\right)^2+4\sqrt{7-x}-4\sqrt{x-1}-\sqrt{\left(7-x\right)\left(x-1\right)}=0\)

\(\Leftrightarrow\left(\sqrt{x-1}\right)\left(\sqrt{x-1}-4\right)+\left(\sqrt{7-x}\right)\left(4-\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-4\right)\left(\sqrt{x-1}-\sqrt{7-x}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=4\\\sqrt{x-1}=\sqrt{7-x}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=17\left(l\right)\\x=4\left(tm\right)\end{cases}}}\)

Answer 2

mà sao bạn k làm giúp mình câu b

Answer 3

bạn ơi mình sai đề câu b

b, \(x^2+2x+4=3\sqrt{x^3+4x}\)