Question

  Làm cho e hết vs ạ

Answer 1

Bài I:

1: \(\sqrt{75}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{2}{2-\sqrt{3}}\)

\(=5\sqrt{3}+\left|\sqrt{3}-1\right|-\dfrac{2\left(2+\sqrt{3}\right)}{4-3}\)

\(=5\sqrt{3}+\sqrt{3}-1-2\left(2+\sqrt{3}\right)\)

\(=6\sqrt{3}-1-4-2\sqrt{3}\)

\(=4\sqrt{3}-5\)

2: 

ĐKXĐ: x>=5

\(\sqrt{x-5}-\sqrt{9x-45}+4=0\)

=>\(\sqrt{x-5}-3\sqrt{x-5}=-4\)

=>\(-2\sqrt{x-5}=-4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

Bài II:

1: Khi x=36 thì \(A=\dfrac{4\cdot\sqrt{36}+1}{\sqrt{36}-2}=\dfrac{4\cdot6+1}{6-2}=\dfrac{25}{4}\)

2: \(B=\dfrac{2}{\sqrt{x}-2}+\dfrac{3\sqrt{x}}{\sqrt{x}+2}-\dfrac{2x}{x-4}\)

\(=\dfrac{2}{\sqrt{x}-2}+\dfrac{3\sqrt{x}}{\sqrt{x}+2}-\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2\left(\sqrt{x}+2\right)+3\sqrt{x}\left(\sqrt{x}-2\right)-2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2\sqrt{x}+4+3x-6\sqrt{x}-2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

3: M=A*B

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{4\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+2}\)

M<3

=>M-3<0

=>\(\dfrac{4\sqrt{x}+1-3\sqrt{x}-6}{\sqrt{x}+2}< 0\)

=>\(\dfrac{\sqrt{x}-5}{\sqrt{x}+2}< 0\)

=>\(\sqrt{x}-5< 0\)

=>\(\sqrt{x}< 5\)

=>\(0< =x< 25\)

mà x là số chính phương và \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

nên \(x\in\left\{0;1;9;16\right\}\)

Answer 2

Bài III:

1: Thay m=1 vào (d), ta được:

\(y=\left(1+2\right)x-3=3x-3\)

Bảng giá trị:

x	0	1
y=3x-3	-3	0

Vẽ đồ thị:

2:

(d): y=(m+2)x-3

(d1): y=-2x+3

Để (d)//(d1) thì \(\left\{{}\begin{matrix}m+2=-2\\3< >-3\left(đúng\right)\end{matrix}\right.\)

=>m+2=-2

=>m=-4

3: Tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\\left(m+2\right)x-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\x\left(m+2\right)=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{3}{m+2}\end{matrix}\right.\)

vậy: \(A\left(\dfrac{3}{m+2};0\right)\)

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=\left(m+2\right)x-3=0\cdot\left(m+2\right)-3=-3\end{matrix}\right.\)

Vậy: B(0;-3)

\(A\left(\dfrac{3}{m+2};0\right);B\left(0;-3\right)\)

\(AB=\sqrt{\left(0-\dfrac{3}{m+2}\right)^2+\left(-3-0\right)^2}\)

\(=\sqrt{\left(\dfrac{3}{m+2}\right)^2+9}\)

Để AB=5 thì \(\sqrt{\left(\dfrac{3}{m+2}\right)^2+9}=5\)

=>\(\left(\dfrac{3}{m+2}\right)^2+9=25\)

=>\(\left(\dfrac{3}{m+2}\right)^2=16\)

=>\(\left[{}\begin{matrix}\dfrac{3}{m+2}=4\\\dfrac{3}{m+2}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m+2=\dfrac{3}{4}\\m+2=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}m=-\dfrac{5}{4}\\m=-\dfrac{11}{4}\end{matrix}\right.\)