Bài này em tham khảo nhé cu :

https://olm.vn/hoi-dap/detail/9206014220.html

Lời giải :

\(\frac{a^2+b^2}{ab}=\frac{\left(a+b\right)^2-2ab}{ab}\)

\(=\frac{1-2ab}{ab}=\frac{1}{ab}-\frac{2ab}{ab}=\frac{1}{ab}-2\)

Ta có : \(ab\le\left(\frac{a+b}{2}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow\frac{1}{ab}-2\ge\frac{1}{\frac{1}{4}}-2=2\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

TL:

\(\frac{a^2+b^2}{ab}=\frac{a^2+2ab+b^2-2ab}{ab}\) 

\(=\frac{\left(a+b\right)^2-2ab}{ab}=\frac{1-2ab}{ab}=\frac{1}{ab}-\frac{2ab}{ab}=\frac{1}{ab}-2\)

mà \(ab\le(\frac{a+b}{2})^2=\frac{1}{4}\) 

\(\frac{1}{ab}-2\ge\frac{\frac{1}{1}}{4}-2=\frac{-7}{4}\)  

\(\Rightarrow ab\ge4\) Dấu "=" xảy ra <=>ab=4(bạn tự tìm a,b nha)

Vậy GTNN của BT=\(\frac{-7}{4}\)

\(B=\frac{x^2+17}{x^2+7}\)

\(\Leftrightarrow Bx^2+7B=x^2+17\)

\(\Leftrightarrow Bx^2+7B-x^2-17=0\)

\(\Leftrightarrow x^2\left(B-1\right)+7B-17=0\)

Để pt có nghiệm thì \(\Delta'\ge0\)

\(\Leftrightarrow0^2-\left(B-1\right)\left(7B-17\right)\ge0\)

\(\Leftrightarrow7B^2-24B+17\le0\)

\(\Leftrightarrow1\le B\le\frac{17}{7}\)

Vậy \(max_B=\frac{17}{7}\Leftrightarrow x=0\)

Phuongdeptrai274:e có cách khác a thử check nha!

\(B=\frac{x^2+17}{x^2+7}\)

\(B=\frac{x^2+7+10}{x^2+7}\)

\(B=1+\frac{10}{x^2+7}\)

\(\Rightarrow B\le1+\frac{10}{0+7}=\frac{17}{7}\)

Dấu "=" xảy ra khi x=0

Lời giải :

\(A=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

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

\(A=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

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

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

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

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

\(A=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

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

\(A=2\left(9-5\right)\)

\(A=8\)

abc+acb+bac+bca+cab+cba

=100a+10b+c+100a+10c+b+100b+10a+c+100b+10c+b+100c+10a+b+100c+10b+a

=(100+10+1).2.a+(100+10+1).2.b+(100+10+1).2.c

=222(a+b+c)

mà 222 chia hết cho 37

suy ra đpcm

Chúc bạn học tốt!

cảm ơn bạn

#)Giải :

Đặt \(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(3A=0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51\)

\(3A=49.50.51=124950\)

\(\Leftrightarrow A=\frac{124950}{3}=41650\)

Mình sửa lại đề vì sai :  1 x 2 + 2 x 3 + ... + 99 x 100

Đặt A = 1 x 2 + 2 x 3 + ... + 99 x 100

  => 3 x A = 3 x (1 x 2 + 2 x 3 + ... + 99 x 100)

  => 3 x A = 1 x 2 x 3 + 2 x 3 x 3 + ... + 99 x 100 x 3

  => 3 x A = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ... + 99 x 100 x (101 - 98)

  => 3 x A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ... + 99 x 100 x 101 - 98 x 99 x 100

  => 3 x A = 99 x 100 x 101

  => 3 x A = 999 900

  => A       = 999 900 : 3

  => A       = 333 300

Vậy 1 x 2 + 2 x 3 + ... + 99 x 100 = 333 300