Question

\(x^2-3x+\sqrt{x^2-3x+2}=10\)\(3\sqrt{x^2-5x+10}=5x-x^2\)\(\left(x+4\right)\left(x-4\right)+3\sqrt{x^2-x+3}+5=0\)\(\left(x-3\right)\left(x+2\right)-2\sqrt{x^2-x+4}+10=0\)

Answer 1

Bài 1:

Đk:\(1\le x\le2\)

\(pt\Leftrightarrow x^2-3x-10=-\sqrt{x^2-3x+2}\)

Đặt \(\sqrt{x^2-3x+2}=t\left(t\ge0\right)\) ta có:

\(t^2-12=-t\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t+4\right)\left(t-3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-4\left(loai\right)\\t=3\end{array}\right.\)

Xét \(t=3\Leftrightarrow x^2-3x+2=3\)

\(\Leftrightarrow x^2-3x-1=0\)

\(\Delta=\left(-3\right)^2-\left[\left(-4\right).\left(1.1\right)\right]=13\)\(\Leftrightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\) (thỏa mãn)

Answer 2

các phần khác tương tự

Answer 3

giai ho minh them cau 4 dc ko

 

Answer 4

Bài 4:

Đk:\(x\in\left\{-\infty;\infty\right\}\)

\(pt\Leftrightarrow x^2-x-6+10=2\sqrt{x^2-x+4}\)

\(\Leftrightarrow x^2-x+4=2\sqrt{x^2-x+4}\)

Đặt \(\sqrt{x^2-x+4}=t\left(t\ge0\right)\) ta được:

\(t^2=2t\Leftrightarrow t^2-2t=0\Leftrightarrow t\left(t-2\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}t=0\\t=2\end{array}\right.\) (thỏa mãn)

Xét \(t=0\Rightarrow\sqrt{x^2-x+4}=0\)

\(\Delta=\left(-1\right)^2-4\left(1.4\right)=-15< 0\) (vô nghiệm)

Xét \(t=2\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x+4=4\)

\(\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\) (thỏa mãn)