đkxđ: \(x\ge0;x\ne4\)

\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)

\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)

\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)

đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

Dat \(\left(a,b,c\right)=\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\left(a,b,c>0,abc=1\right)\)

Ta co \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\frac{3}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}\left(1\right)\)

BDT phu \(1+\frac{3}{ab+bc+ca}\ge\frac{6}{a+b+c}\left(2\right)\)

Do (1) nen (2) tuong duong voi

\(1+\frac{9}{\left(a+b+c\right)^2}\ge\frac{6}{a+b+c}\Leftrightarrow\left(1-\frac{3}{a+b+c}\right)^2\ge0\left(dung\right)\)

Suy ra (2) duoc chung minh

Do \(abc=1\Rightarrow\hept{\begin{cases}ab=\frac{1}{xy}=\frac{xyz}{xy}=z\\bc=x\\ca=y\end{cases}}\)

nen (2) tuong duong \(1+\frac{3}{x+y+z}\ge\frac{6}{xy+yz+zx}\)

=> \(\frac{1}{x+y+z}\ge\frac{1}{3}\left(\frac{6}{x+y+z}-1\right)=\frac{2}{x+y+z}-\frac{1}{3}\)

Suy ra \(P\ge\frac{2}{x+y+z}-\frac{1}{3}-\frac{2}{x+y+z}=-\frac{1}{3}\)

Dau = xay ra khi x=y=z=1

\(\left(\sqrt{2x+1}+\sqrt{2x+16}\right)^2=\left(\sqrt{2x+4}+\sqrt{2x+9}\right)^2\)          

\(2x+1+2x+16+2\sqrt{\left(2x+1\right)\left(2x+16\right)}=2x+4+2x+9+2\sqrt{\left(2x+4\right)\left(2x+9\right)}\)     

\(4x+17+2\sqrt{4x^2+34x+16}=4x+13+2\sqrt{4x^2+26x+36}\) 

\(2+\sqrt{4x^2+34x+16}=\sqrt{4x^2+26x+36}\)  

\(4+4x^2+34x+16+4\sqrt{4x^2+34x+16}=4x^2+26x+36\)      

\(4\sqrt{4x^2+34x+16}=-8x+16\)     

\(\sqrt{4x^2+34x+16}=-2x+4\)         

\(\hept{\begin{cases}-2x+4\ge0\\4x^2+34x+16=\left(-2x+4\right)^2\end{cases}}\)                            

\(\hept{\begin{cases}-2x\ge-4\\4x^2+34x+16=4x^2-16x+16\end{cases}}\)      

\(\hept{\begin{cases}x\le2\\50x=0\end{cases}}\)                                                                                                                   

\(\hept{\begin{cases}x\le2\\x=0\end{cases}}\) 

\(\Rightarrow x=0\)     

Câu 1:

G/s \(\sqrt{7}\) là số hữu tỉ có thể viết dưới dạng phân số tối giản \(\frac{a}{b}\) \(\left(a,b\inℤ\right)\)

=> \(\frac{a}{b}=\sqrt{7}\)

<=> \(\left(\frac{a}{b}\right)^2=7\)

=> \(a^2=7b^2\)

=> \(a^2⋮b^2\) , mà theo đề bài phân số tối giản

=> a không chia hết cho b => a² không chia hết cho b² 

=> vô lý

=> \(\sqrt{7}\) là số vô tỉ

Câu 2:

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=\left(a^2c^2+a^2d^2\right)+\left(b^2c^2+b^2d^2\right)\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

b) Ta có: \(\left(ac+bd\right)^2=a^2c^2+2abcd+b^2d^2\)

\(=a^2c^2+2\sqrt{a^2d^2.b^2c^2}+b^2d^2\)

\(\le a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\) ( bất đẳng thức Cauchy )

Dấu "=" xảy ra khi: \(ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

\(2\left(x^2+2\right)=5\sqrt{x^3+1}\left(đk:x\ge-1\right)\)

\(\Leftrightarrow2\left[\left(x^2-x+1\right)+\left(x+1\right)\right]=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2+1}=a\left(a\ge0\right)\\\sqrt{x^2-x+1}=b\left(b>0\right)\end{cases}}\)

Tìm được \(\orbr{\begin{cases}a=2b\\b=2a\end{cases}}\)

TH1: a=2b => phương trình vô nghiệm

TH2: b=2a ta được \(x_1=\frac{5+\sqrt{37}}{2};x_2=\frac{5-\sqrt{37}}{2}\left(tmđk\right)\)

\(\sqrt{8+\sqrt{15}}\)                                      

=\(\sqrt{\frac{15}{2}+2\cdot\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}\)         

=\(\sqrt{\left(\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)                                                              

=\(|\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}|\)            

=\(\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}\)                  

\(\sqrt{9-\sqrt{77}}\)     

=\(\sqrt{\frac{11}{2}-2\cdot\sqrt{\frac{11}{2}}\cdot\sqrt{\frac{7}{2}}+\frac{7}{2}}\)      

=\(\sqrt{\left(\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}\right)^2}\)     

=\(|\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}|\)         

=\(\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}\)       

\(\sqrt{10+\sqrt{99}}\)                         

=\(\sqrt{\frac{11}{2}+2\cdot\sqrt{\frac{11}{2}}\cdot\sqrt{\frac{9}{2}}+\frac{9}{2}}\)      

=\(\sqrt{\left(\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}\right)^2}\)                          

=\(|\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}|\)                                         

=\(\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}\)                                                        

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

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

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

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