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
\(ĐK:x\ge1\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)+2.2.\sqrt{x-1}+4}+\sqrt{\left(x-1\right)-2.\sqrt{x-1}.3+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=5\)
\(\Leftrightarrow\sqrt{x-1}+2+\left|\sqrt{x-1}-3\right|=5\)
TH1: \(\sqrt{x-1}-3\ge0\Leftrightarrow x\ge10\)
\(pt\Leftrightarrow\sqrt{x-1}+2+\sqrt{x-1}-3=5\)
\(\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(tm\right)\)
TH2: \(\sqrt{x-1}-3< 0\Leftrightarrow1\le x< 10\)
\(pt\Leftrightarrow\sqrt{x-1}+2-\sqrt{x-1}+3=5\)
\(\Leftrightarrow5=5\) (Đúng với mọi x)
Vậy phương trình có nghiệm là: \(x\in\left[1;10\right]\)
