tách hình ra chứ để z chả ai rảnh làm hết đâu :vv
l: Ta có: \(\sqrt{x^2-10x+25}=5-x\)
\(\Leftrightarrow\left|x-5\right|=5-x\)
\(\Leftrightarrow x-5\le0\)
hay \(x\le5\)
\(a,ĐK:x\ge5\\ PT\Leftrightarrow2\sqrt{x-5}+3\cdot\dfrac{1}{3}\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=4\\ \Leftrightarrow4\sqrt{x-5}=4\\ \Leftrightarrow x-5=1\Leftrightarrow x=6\left(tm\right)\\ b,ĐK:x\ge1\\ PT\Leftrightarrow\dfrac{2}{3}\cdot3\sqrt{x-1}-\dfrac{1}{4}\cdot4\sqrt{x-1}+27\cdot\dfrac{1}{9}\sqrt{x-1}=4\\ \Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\\ c,ĐK:2\le x\le8\\ PT\Leftrightarrow x-2=x^2-16x+64\\ \Leftrightarrow x^2-17x+66=0\\ \Leftrightarrow\left(x-11\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=11\left(ktm\right)\\x=6\left(tm\right)\end{matrix}\right.\Leftrightarrow x=6\)
\(d,ĐK:x\ge8\\ PT\Leftrightarrow x-2=x^2-16x+6x\left(làm.như.câu.c\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=11\left(tm\right)\\x=6\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=11\)
