Câu 3:
a: Thay \(m=-\frac32\) vào phương trình, ta được:
\(x^2-2\left(-\frac32+2\right)x+\frac{-3}{2}+1=0\)
=>\(x^2-2\cdot\frac12x-\frac12=0\)
=>\(x^2-x-\frac12=0\)
=>\(x^2-x+\frac14-\frac34=0\)
=>\(\left(x-\frac12\right)^2=\frac34\)
=>\(x-\frac12=\pm\frac{\sqrt3}{2}\)
=>\(x=\frac{1\pm\sqrt3}{2}\)
b: \(\Delta=\left\lbrack-2\left(m+2\right)\right\rbrack^2-4\left(m+1\right)\)
\(=4\left(m^2+4m+4\right)-4\left(m+1\right)\)
\(=4\left(m^2+4m+4-m-1\right)=4\left(m^2+3m+3\right)\)
\(=4\left(m^2+3m+\frac94+\frac34\right)\)
\(=4\left(m+\frac32\right)^2+3\ge3>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
c: Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2\left(m+2\right)\\ x_1x_2=\frac{c}{a}=m+1\end{cases}\)
\(x_1^2+x_2^2=8\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=8\)
=>\(\left(2m+4\right)^2-2\left(m+1\right)=8\)
=>\(4m^2+16m+16-2m-2-8=0\)
=>\(4m^2+14m+6=0\)
=>\(2m^2+7m+3=0\)
=>\(2m^2+6m+m+3=0\)
=>(m+3)(2m+1)=0
=>\(\left[\begin{array}{l}m+3=0\\ 2m+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=-3\\ m=-\frac12\end{array}\right.\)
Câu 2:
a:
ĐKXĐ: x∈R
\(\sqrt{x^2-4x+4}+x=8\)
=>\(\left|x-2\right|=8-x\)
=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\le8\\ \left(x-8\right)^2-\left(x-2\right)^2=0\end{cases}\)
=>\(\begin{cases}x\le8\\ \left(x-8-x+2\right)\left(x-8+x-2\right)=0\end{cases}\Rightarrow\begin{cases}x\le8\\ -6\left(2x-10\right)=0\end{cases}\)
=>\(\begin{cases}x\le8\\ 2x-10=0\end{cases}\Rightarrow\begin{cases}x\le8\\ x=5\end{cases}\)
=>x=5
b: \(\begin{cases}x+y=4\\ 2x-y=-7\end{cases}\Rightarrow\begin{cases}x+y+2x-y=4-7=-3\\ x+y=4\end{cases}\)
=>\(\begin{cases}3x=-3\\ x+y=4\end{cases}\Rightarrow\begin{cases}x=-1\\ y=4-x=4-\left(-1\right)=5\end{cases}\)
Câu 1:
a: \(A=\sqrt5\left(\sqrt{20}-3\right)+\sqrt{45}=\sqrt{100}-3\sqrt5+3\sqrt5=\sqrt{100}=10\)
b: \(\sqrt{24+16\sqrt2}-\sqrt{24-16\sqrt2}\)
\(=\sqrt{8\left(3+2\sqrt2\right)}-\sqrt{8\left(3-2\sqrt2\right)}\)
\(=\sqrt8\cdot\left(\sqrt{3+2\sqrt2}-\sqrt{3-2\sqrt2}\right)\)
\(=2\sqrt2\left(\sqrt{\left(\sqrt2+1\right)^2}-\sqrt{\left(\sqrt2-1\right)^2}\right)=2\sqrt2\left(\sqrt2+1-\sqrt2+1\right)=2\sqrt2\cdot2=4\sqrt2\)
c: ĐKXĐ: x>=-1/2
\(\sqrt{2x+1}\le5\)
=>2x+1<=25
=>2x<=24
=>x<=12
=>\(-\frac12\le x<=12\)
