10) Ta có: x2 - 16x + 60 = 0
=> x2 - 2x8 + 82 - 82 + 60 = 0
=> (x2 - 2x8 + 82) - (64 - 60) = 0
=> (x - 8)2 - 4 = 0
=> (x - 8)2 = 4
=> x - 8 thuộc {-2; 2}
=> x thuộc {6; 10}
Vậy x thuộc {6; 10}
11) Ta có : x2 - 6x = -13
=> x2 - 6x + 13 = 0
=> x2 - 2x3 + 32 - 32 + 13 = 0
=> (x2 - 2x3 + 32) + (13 - 9) = 0
=> (x - 3)2 + 4 = 0
=> (x - 3)2 = -4
mà (x - 3)2 > với mọi x
=> k tìm đc số x thỏa mãn
Vậy..
Chúc bạn học tốt!
10)\(x^2-16x+60=0\)
\(\Leftrightarrow x^2-10x-6x+60=0\)
\(\Leftrightarrow x\left(x-10\right)-6\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-6=0\\x-10=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=10\end{array}\right.\)
\(10,x^2-16x+60=0\)
\(\Leftrightarrow\left(x^2-2.x.8+8^2\right)-8^2+60=0\)
\(\Leftrightarrow\left(x-8\right)^2-4=0\)
\(\Leftrightarrow\left(x-8\right)^2-2^2=0\)
\(\Leftrightarrow\left(x-8-2\right)\left(x-8+2\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-10=0\\x-6=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\x=6\end{array}\right.\)
\(\text{Vậy x=10 hoặc x=6 }\)
