a) điều kiện \(x\ge1\)
ta có \(A=\sqrt{25x-25}-\sqrt{16x-16}-\sqrt{4x+4}\)
\(\Leftrightarrow A=\sqrt{25\left(x-1\right)}-\sqrt{16\left(x-1\right)}-\sqrt{4\left(x+1\right)}\)
\(\Leftrightarrow A=5\sqrt{x-1}-4\sqrt{\left(x-1\right)}-2\sqrt{x+1}=\sqrt{x-1}-2\sqrt{x+1}\)
b) để \(A=10\Leftrightarrow\sqrt{x-1}-2\sqrt{x+1}=10\)
\(\Leftrightarrow x-1+4x+4-4\sqrt{\left(x^2-1\right)}=100\)
\(\Leftrightarrow5x-97=4\sqrt{x^2-1}\Leftrightarrow25x^2-970x+9409=16x^2-16\)
\(\Leftrightarrow9x^2-970x+9425\Rightarrow x\)
nhớ điều kiện nha :)
a) A=\(\sqrt{25\left(x-1\right)}-\sqrt{16\left(x-1\right)}+\sqrt{4\left(x-1\right)}\)
A=\(\sqrt{x-1}\left(\sqrt{25}-\sqrt{16}+\sqrt{4}\right)\)
A=\(3\sqrt{x-1}\)
b) Ta có A=10 => \(3\sqrt{x-1}\)=10=>x-1=\(\dfrac{100}{9}\)=>x=\(\dfrac{109}{9}\)