Câu 9 .2)

Có: \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=1\)

\(\Leftrightarrow\) \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=1-xy\)

\(\Leftrightarrow\) \(\left(1+x^2\right)\left(1+y^2\right)=\left(1-xy\right)^2\)

\(\Leftrightarrow\) \(1+x^2+y^2+x^2y^2=1-2xy+x^2y^2\)

\(\Leftrightarrow\) \(x^2+y^2=-2xy\)

\(\Leftrightarrow\) \(x^2+y^2+2xy=0\)

\(\Leftrightarrow\) \(\left(x+y\right)^2=0\)

\(\Leftrightarrow\) \(x+y=0\Leftrightarrow y=-x\)

\(\Rightarrow x\sqrt{1+y^2}+y\sqrt{1+x^2}=x\sqrt{1+\left(-x\right)^2}+\left(-x\right)\sqrt{1+x^2}\)

\(=x\sqrt{1+x^2}-x\sqrt{1+x^2}=0\left(đpcm\right)\)

Cơ quan, công chức nhà nước có thẩm quyền ban hành các quyết định trong quản lý, điều hành là biểu hiện của hình thức thực hiện pháp luật nào dưới đây?

A. Sử dụng pháp luật.

B. Thi hành pháp luật.

C. Tuân thủ pháp luật.

D. Áp dụng pháp luật.

Tuân thủ pháp luật là hình thức, thực hiện pháp luật trong đó các cá nhân, tổ chức không làm những điều mà pháp luật

A. cho phép làm.

B. quy định cấm.

C. quy định phải làm.

D. không bắt buộc.

1a)\(\sqrt{3-2\sqrt{2}}=\sqrt{2-2.\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

b)\(\sqrt{28+10\sqrt{3}}=\sqrt{25+2.5.\sqrt{3}+3}=\sqrt{\left(5+\sqrt{3}\right)^2}=\left|5+\sqrt{3}\right|=5+\sqrt{3}\)

c)\(\sqrt{14+6\sqrt{5}}=\sqrt{9+2.3.\sqrt{5}+5}=\sqrt{\left(3+\sqrt{5}\right)^2}=\left|3+\sqrt{5}\right|=3+\sqrt{5}\)

d)\(\sqrt{13-4\sqrt{3}}=\sqrt{13-2.\sqrt{12}}=\sqrt{12-2.\sqrt{12}.1+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)

2)a)\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2.\sqrt{7}.1+1}-\sqrt{7+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|=\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)

b)\(\sqrt{18+8\sqrt{2}}-\sqrt{18-8\sqrt{2}}\)

\(=\sqrt{16+2.4.\sqrt{2}+2}-\sqrt{16-2.4.\sqrt{2}+2}\)

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

\(=\left|4+\sqrt{2}\right|-\left|4-\sqrt{2}\right|=4+\sqrt{2}-\left(4-\sqrt{2}\right)=2\sqrt{2}\)

\(y=2x^2-\left(m+3\right)x+m=0\)

\(\Delta=\left[-\left(m+3\right)\right]^2-4.2.m=m^2+6m+9-8m=m^2-2m+9>0\forall m\)

\(\sqrt{\Delta}=\sqrt{m^2-2m+9}\)

\(\Rightarrow x_1=\dfrac{m+3+\sqrt{m^2-2m+9}}{4}\);\(x_2=\dfrac{m+3-\sqrt{m^2-2m+9}}{4}\)

\(\Rightarrow P=\left|x_1-x_2\right|=\left|\dfrac{m+3+\sqrt{m^2-2m+9}}{4}-\dfrac{m+3-\sqrt{m^2-2m+9}}{4}\right|\)

\(P=\left|\dfrac{2\sqrt{m^2-2m+9}}{4}\right|=\dfrac{\sqrt{m^2-2m+9}}{2}=\dfrac{\sqrt{\left(m-1\right)^2+8}}{2}\)

vì \(\left(m-1\right)^2\ge0\Rightarrow\left(m-1\right)^2+8\ge8\Rightarrow\sqrt{\left(m-1\right)^2+8}\ge\sqrt{8}=2\sqrt{2}\)

\(\Rightarrow P=\dfrac{\sqrt{\left(m-1\right)^2+8}}{2}\ge\dfrac{2\sqrt{2}}{2}=\sqrt{2}\)

vậy minP=\(\sqrt{2}\) khi m-1=0 <=> m=1

a) \(A=\sqrt{\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)}\)

\(A=\sqrt{1+x^2+y^2+\left(xy\right)^2+4xy+2\left(x+y+x^2y+xy^2\right)}\)

\(A=\sqrt{1+x^2+y^2+\left(xy\right)^2+4xy+2x+2y+2x^2y+2xy^2}\)

\(A=\sqrt{1^2+x^2+y^2+\left(xy\right)^2+2.x.1+2.y.1+2xy+2.xy+2x.xy+2y.xy}\)

\(A=\sqrt{\left(1+x+y+xy\right)^2}\)

\(A=\left|1+x+y+xy\right|=1+x+y+xy\)(vì x,y nguyên dương)

b)A=1+x+y+xy=6

(x+1)+y(x+1)=6

<=> (x+1)(y+1)=6=1.6=2.3=3.2=6.1

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=1\\y+1=6\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=2\\y+1=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=3\\y+1=2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=6\\y+1=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\end{matrix}\right.\)

vạy (x;y)=(0;5);(1;2);(2;1);(5;0)

tam giác ABC vuông cân tại C => CA=CB

D là trung điểm của CA

=> \(DC=\dfrac{1}{2}AC=\dfrac{1}{2}CB\)

Có góc DCM = góc CBD ( cùng phụ với góc BDC)

=> \(tanDCM=\tan CBD=\dfrac{CD}{CB}=\dfrac{\dfrac{1}{2}CB}{CB}=\dfrac{1}{2}\)

Vậy \(\tan ACM=\dfrac{1}{2}\)

ta có:\(\cot\alpha=2\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\Rightarrow\dfrac{\cos^2\alpha}{\sin^2\alpha}=4\Rightarrow\cos^2\alpha=4\sin^2\alpha\)

\(\Rightarrow\dfrac{sin^2\alpha-\cos^2\alpha}{sin^2\alpha+\cos^2\alpha}=\dfrac{sin^2\alpha-4sin^2\alpha}{sin^2\alpha+4sin^2\alpha}=\dfrac{-3sin^2\alpha}{5sin^2\alpha}=-\dfrac{3}{5}\left(ĐPCM\right)\)

d) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{5-2.2\sqrt{5}+4}-\sqrt{5+2.2\sqrt{5}+4}\)

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

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

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

g)\(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

\(=\dfrac{\sqrt{3}+\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}}{\sqrt{2}+\sqrt{5+2.\sqrt{5}.1+1}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}}\)

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

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

\(=\dfrac{3}{1}=3\)

\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3+\sqrt{6}}-\sqrt{6}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3-\sqrt{6}\right)}{3}-\sqrt{6}\)

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

\(=3\sqrt{6}-3+2\sqrt{6}+4-12+4\sqrt{6}-\sqrt{6}\)

\(=8\sqrt{6}-11\)