Gọi G là trọng tâm tam giác ABC

Vì là trung tuyến \(\Rightarrow\hept{\begin{cases}BN=\frac{3}{2}BG\\CP=\frac{3}{2}CG\end{cases}}\)

\(\Rightarrow BN+CP=\frac{3}{2}\left(BG+CG\right)\)

Mà theo bđt trong tam giác cho tam giác BGC thì \(BG+GC>BC\)

\(\Rightarrow BN+CP>\frac{3}{2}BC\)

\(\left(4+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)

\(x^3-2x^2-5x+6\)

\(=\left(x^3-4x^2+3x\right)+\left(2x^2-8x+6\right)\)

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

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

1. \(x^3-2x-5x+6\)

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

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

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

2. \(x^3-7x^2+15x-9\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\)

\(\Leftrightarrow x^2\left(x-1\right)-6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-6x+9\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x\left(x-3\right)-3\left(x-3\right)\right)\)

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

Em thử ạ. Bài dài quá em chẳng biết có tính sai chỗ nào hay không nữa ;(

Từ giả thiết ta có: 

\(\hept{\begin{cases}x+y=-\frac{2}{3}\left(z+1\right)\\xy=-\frac{1}{3}\end{cases}}\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=\frac{4}{9}\left(z+1\right)^2+\frac{2}{3}\)

Và \(\left(x-y\right)^2=\left(x+y\right)^2-4xy=\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}\)

Ta có: \(A=\frac{\left(x-y\right)\left(x^2+xy+y^2\right)+\left(z+1\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{\left(x-y\right)^3}\)

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

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

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

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-\frac{2}{9}\left(z+1\right)^2+\frac{1}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-\frac{2}{9}\left(z+1\right)^2+\frac{1}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

Ơ....hình như em tính sai chỗ nào rồi:(

Nguyễn Khang 

\(A=\frac{\left(x^2+y^2-\frac{1}{3}+\left(z+1\right)\left(x+y\right)-1\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2-1\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\) ( như này mới đúng, e thiếu -1 ở tử ) 

\(=\frac{\frac{-2}{9}\left(z+1\right)^2-\frac{2}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=-\frac{1}{2}.\frac{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-1}{2}\)

=11112324

Học tốt

1+11111111+1212=1112324

nhớ k nhưng ko làm bạn gái

các bạn giải hộ mình cái

Xét tam giác ABF có  góc ABE là góc ngoài tại định B

=> \(\widehat{ABE}=\widehat{BAF}+\widehat{AFB}\) 

Xét tam giác FBC có: \(\widehat{CBE}\)là góc ngoài tại định B

=> \(\widehat{CBE}=\widehat{BFC}+\widehat{CFB}\)

Cộng vế theo vế ta có:

\(\widehat{CBE}+\widehat{ABE}=\widehat{BFC}+\widehat{CFB}+\widehat{AFB}+\widehat{BAF}>\widehat{AFB}+\widehat{CFB}\)

=> \(\widehat{ABC}>\widehat{AFC}\)

=> \(\widehat{AFC}< 90^o\)

hay AFC là góc nhọn

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

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

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

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

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

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

1a) \(\frac{5}{1,2}=\frac{-2,5}{x}\)

\(\Leftrightarrow5x=-3\)

\(\Leftrightarrow x=\frac{-3}{5}\)

 b) \(\frac{3,2+\left(-0,4\right)}{-x-3,6}=\frac{-0,75}{1,5}\)

\(\Leftrightarrow\frac{2,8}{-x-3,6}=\frac{-0,75}{1,5}\)

\(\Leftrightarrow4,2=0,75x+2,7\)

\(\Leftrightarrow0,75x=1,5\)

\(\Leftrightarrow x=2\)

2) \(\frac{1}{3}.\frac{5}{7}=\frac{2}{7}.\frac{5}{6}\)

Tỉ lệ thức lập được \(\frac{5}{21}=\frac{10}{42}\)

1. 15^x=225

   <=>15^x=15^2

    =>x=2

    vậy....

2. x^3=27

   <=>x^3=3^3

   =>x=3

vậy......

3. 4.2^x-3=125

   4.2^x=128

   2^x=32

   2^x=2^5

   =>x=5

vậy.....

\(15^x=225\)

\(\Leftrightarrow15^x=15^2\)

\(\Leftrightarrow x=2\)

\(x^3=27\)

\(\Leftrightarrow x^3=3^3\)

\(\Leftrightarrow x=3\)

\(4.2^x-3=125\)

\(\Leftrightarrow4.2^x=125+3\)

\(\Leftrightarrow4.2^x=128\)

\(\Leftrightarrow2^x=128:4\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

~ Hok tốt ~