\(\Leftrightarrow x=0\Leftrightarrow\left\{{}\begin{matrix}5-m=y\\3+m=y\end{matrix}\right.\Leftrightarrow y=4\Leftrightarrow5-m=4\Leftrightarrow m=1\)

Áp dụng BĐT AM-GM ta có:

\(\left(a+1\right)^2+b^2+1=a^2+2a+1+b^2+1=\left(a^2+b^2\right)+2a+2\ge2\left(ab+a+1\right)\)

\(\Rightarrow\frac{1}{\left(a+1\right)^2+b^2+1}\le\frac{1}{2\left(ab+a+1\right)}\)(1)

\(\left(b+1\right)^2+c^2+1=b^2+2b+1+c^2+1=\left(b^2+c^2\right)+2b+2\ge2\left(bc+b+1\right)\)

\(\Rightarrow\frac{1}{\left(b+1\right)^2+c^2+1}\le\frac{1}{2\left(bc+b+1\right)}\)(2)

\(\left(c+1\right)^2+a^2+1=c^2+2c+1+a^2+1=\left(c^2+a^2\right)+2c+2\ge2\left(ca+c+1\right)\)

\(\Rightarrow\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2\left(ca+c+1\right)}\)(3)

Cộng vế theo vế của (1) ; (2) ; (3) ta được:

\(\frac{1}{\left(a+1\right)^2+b^2+1}+\frac{1}{\left(b+1\right)^2+c^2+1}+\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{1}{2}\)Dấu "=" xảy ra \(\Leftrightarrow a=b=b=1\)

BÀI 3:

bài 4:

\(A=\frac{x}{\sqrt{x}-1}+\frac{\sqrt{x}-2x}{x-\sqrt{x}}\)

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

bn làm tắt à

lên mạng mà tìm

a: BC=5cm

\(\widehat{B}=37^0\)

\(\widehat{C}=53^0\)

bạn giải chi tiết ra luôn đc ko ạ

1B  2B  3D  4C  5B  6B  7A  8D  9A  10C  11D  12A

1B

2B

3D

4C

5B

6B

7A

8D

9B

10C

11D

12A

7:

a: ĐKXĐ: x>=0; x<>1

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

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

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

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

b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)

c: |D|=1/3

=>D=-1/3 hoặc D=1/3

=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}+1=3\)

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

=>x=4

6:

a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)

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

b: C<-1

=>C+1<0

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

=>\(-\sqrt{x}+4< 0\)

=>\(-\sqrt{x}< -4\)

=>\(\sqrt{x}>4\)

=>x>16

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Để `C < -1` Ta có :

 \(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)