mình giúp bài 3 cho 

\(\sqrt{25x-125}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=6\left(ĐKXĐ:x\ge5\right)\)

\(< =>\sqrt{25\left(x-5\right)}-3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=6\)

\(< =>\sqrt{25}.\sqrt{x-5}-3\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=6\)

\(< =>5.\sqrt{x-5}-3.\frac{\sqrt{x-5}}{3}-\frac{1}{3}.3.\sqrt{x-5}=6\)

\(< =>5.\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\)

\(< =>3\sqrt{x-5}=6< =>\sqrt{x-5}=2\)

\(< =>x-5=4< =>x=4+5=9\left(tmđk\right)\)

ĐẶT:    \(a=\sqrt[3]{\sqrt{2}-1}\)

=>   \(a^3=\sqrt{2}-1\)

=>   \(x=a-\frac{1}{a}\)

=>   \(x^3=a^3-\frac{1}{a^3}-3a+\frac{3}{a}\)

<=>   \(x^3=\sqrt{2}-1-\frac{1}{\sqrt{2}-1}-3\left(a-\frac{1}{a}\right)\)

<=>   \(x^3=\frac{\left(\sqrt{2}-1\right)^2-1}{\sqrt{2}-1}-3x\)

<=>   \(x^3+3x=\frac{3-2\sqrt{2}-1}{\sqrt{2}-1}\)

<=>   \(x^3+3x=\frac{2-2\sqrt{2}}{\sqrt{2}-1}\)

<=>   \(x^3+3x=\frac{2\left(1-\sqrt{2}\right)}{\sqrt{2}-1}\)

<=>   \(x^3+3x=-2\)

<=>   \(x^3+3x+2=0\Rightarrow P=0\)

VẬY    \(P=0\)

Đặt \(a=\sqrt[3]{\sqrt{2}-1};b=\frac{1}{\sqrt[3]{\sqrt{2}-1}}\Rightarrow\hept{\begin{cases}x=a-b\\ab=1\end{cases}}\)

Xét \(x^3=\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\)

\(x^3=\left(\sqrt{2}-1\right)-\frac{1}{\sqrt{2}-1}-3x\)

\(\Leftrightarrow x^3=-2-3x\Leftrightarrow x^3+3x+2=0\)

Vậy P=0

a) 

\(1+tan^2a=\frac{1}{cos^2a}\)           

\(1+3^2=\frac{1}{cos^2a}\)  

\(10=\frac{1}{cos^2a}\)     

\(cos^2a=\frac{1}{10}\) 

\(cosa=\pm\sqrt{\frac{1}{10}}=\pm\frac{1}{\sqrt{10}}\)    

\(sin^2a+cos^2a=1\)                                                             

\(sin^2a+\frac{1}{10}=1\)   

\(sin^2a=\frac{9}{10}\)     

\(sina=\pm\sqrt{\frac{9}{10}}=\pm\frac{3}{\sqrt{10}}\)   

Vì tan = 3 nên M có 2 trường hợp : 

TH1 : 

sin và cos cùng dương 

\(\Rightarrow M=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}}\)     

\(=\frac{\frac{4}{\sqrt{10}}}{-\frac{2}{\sqrt{10}}}\)                        

= -2 

TH2 : 

Cả sin và cos cùng âm 

\(\Rightarrow M=\frac{-\frac{1}{\sqrt{10}}+\left(-\frac{3}{\sqrt{10}}\right)}{-\frac{1}{\sqrt{10}}-\left(-\frac{3}{\sqrt{10}}\right)}\)            

=\(\frac{-\frac{4}{\sqrt{10}}}{\frac{2}{\sqrt{10}}}\)                 

= -2 

b) 

\(B=\frac{sin15+cos15}{cos15}-cot75\)         

=\(\frac{sin15}{cos15}+\frac{cos15}{cos15}-cot75\)          

=\(tan15+1-cot75\)     

=\(cot75+1-cot75\)    

= 1 

a) đkxđ: \(a>0;a\ne1\)

Ta có:

\(P=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(1-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(P=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}-1}{\sqrt{a}}.\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\frac{2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{2\sqrt{a}\left(\sqrt{a}+1\right)+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{2a+2\sqrt{a}+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

\(P=\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

b) \(P=7\)

\(\Leftrightarrow\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}=7\)

\(\Leftrightarrow4a+2\sqrt{a}+2=7a+7\sqrt{a}\)

\(\Leftrightarrow3a+5\sqrt{a}-2=0\)

\(\Leftrightarrow\left(3a-\sqrt{a}\right)+\left(6\sqrt{a}-2\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\sqrt{a}+2\left(3\sqrt{a}-1\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\left(\sqrt{a}+2\right)=0\)

Mà \(\sqrt{a}+2\ge2\left(\forall a\right)\)

\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)

\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

a) đkxđ: \(x\ne\pm1\)

Ta có:

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

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

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

\(P=\frac{4x}{\left(x-1\right)\left(x+1\right)}\div\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(P=\frac{4x}{\left(x+1\right)^2}\)

b) Ta có: \(x=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

=> \(P=\frac{4\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1+1\right)^2}=\frac{4\sqrt{3}+4}{\left(\sqrt{3}+2\right)^2}=\frac{4+4\sqrt{3}}{7+4\sqrt{3}}\)

c) \(P=-3\)

\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=-3\)

\(\Leftrightarrow-3\left(x^2+2x+1\right)=4x\)

\(\Leftrightarrow-3x^2-6x-3=4x\)

\(\Leftrightarrow3x^2+10x+3=0\)

\(\Leftrightarrow\left(3x^2+x\right)+\left(9x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=-3\end{cases}}\)

Giả sử phản chứng n ko chia hết cho 5 

=> n có dạng là 5a + 1; 5b + 2; 5c + 3; 5d + 4

TH1:   n = 5a + 1

=>   \(n^2=\left(5a+1\right)^2=25a^2+10a+1\)     ko chia hết cho 5

TH2:   n = 5b + 2

=>    \(n^2=\left(5b+2\right)^2=25b^2+20b+4\)    ko chia hết cho 5

TH3:   n = 5c + 3

=>   \(n^2=\left(5c+3\right)^2=25c^2+30c+9\)     ko chia hết cho 5

TH4:   n = 5d + 4

=>   \(n^2=\left(5d+4\right)^2=25d^2+40d+16\)  ko chia hết cho 5

VẬY QUA 4 TRƯỜNG HỢP THÌ TA THẤY ĐIỀU GIẢ SỬ LÀ SAI

=>    ĐIỀU PHẢI CHỨNG MINH:     \(n^2⋮5\Rightarrow n⋮5\)

Giả sử n² chia hết cho 5 và n không chia hết cho 5.

Nếu n=5k\(\pm\)1 \(\left(k\inℕ\right)\)thì \(n^2=25k^2\pm10k+1=5\left(5k^2\pm2k\right)+1⋮̸5\)

Nếu \(n=5k\pm2\left(k\inℕ\right)\)thì \(n^2=25k^2\pm20k+4=5\left(5k^2\pm4k\right)+4⋮̸5\)

Điều này mâu thuẫn với giả thiết n² chia hết cho 5

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-\left(a^3+b^3+c^3\right)\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

VẬY \(\left(a+b+c\right)^3-\left(a^3+b^3+c^3\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

VẬY TA ĐÃ PHÂN TÍCH NHÂN TỬ XONG !!!!