Ta có x-6=7-6=1

=>A=(x-4)^(x-5)=(7-4)^(7-5)=3²=9

Vậy giá trị của A tại x=7 là 9

Thay x = 7 vào p, ta đc:

\(p=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{7+5}}}}\)

     \(=3^{2^{1^{13^{12}}}}=9\)

x = 7 => x - 6 = 1

=> \(\left(x-6\right)^{\left(x+6\right)^{\left(x+5\right)}}=1^{\left(x+6\right)^{\left(x+5\right)}}=1\)

\(\Rightarrow P=\left(x-4\right)^{\left(x-5\right)^1}=\left(7-4\right)^{\left(7-5\right)}=3^2=9\)

Vì | x-1| ; |x+2|; |x-3| ; |x+4| ; |x-5|; |x+6| ; |x-7| ; |x+8| ; |x-9| luôn luôn < hoặc = 0

vì vậy min của T =0

\(T=|x-1|+|x+2|+|x-3|+|x+4|+|x-5|+|x+6|+|x-7|+|x+8|+|x-9|\)

\(\Rightarrow T=|x-1|+|x+2|+|3-x|+|x+4|+|5-x|+|x+6|+|7-x|+|x+8|+|9-x|\)

\(\Rightarrow T\ge|x-1+x+2+3-x+x+4+5-x+x+6+7-x+x+8+9-x|\)

\(\Rightarrow T\ge|43|\)

\(\Rightarrow T\ge43\)

Vậy \(Min_T=43\)

ko nhìn thấy bợn ưi

1. Ta có: \(\sqrt{23}+\sqrt{15}< \sqrt{25}+\sqrt{16}=5+4=9\)

mà \(\sqrt{83}>\sqrt{81}=9\)

\(\Rightarrow\sqrt{23}+\sqrt{15}< \sqrt{83}\)

A=(3x+7)(2x+3)-(3x-5)(2x+11)  =6x²+9x+14x+21-6x²-33x+10x+55          =(6x²-6x²)+(9x+14x-33x+10x)+(21+55)  =76

\(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(\Leftrightarrow A=6x^2+14x+9x+21-\left(6x^2-10x+33x-55\right)\)

\(\Leftrightarrow A=6x^2+23x+21-\left(6x^2+23x-55\right)\)

\(\Leftrightarrow A=6x^2+23x+21-6x^2-23x+55\)

\(\Leftrightarrow A=76\)

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

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

\(\Leftrightarrow B=x^3+x^2-x^2-x-x-1-x^3+x^2-x^2+x-x+1\)

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

\(\Leftrightarrow B=-2x\)