\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x=100\) vì \(\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)\ne0\)

\(\Leftrightarrow S=\left\{100\right\}\)

a,=(x\(^2\)-6x+9)+10-9

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

Mà(x-3)\(^2\)\(\ge\)0

nên (x-3)\(^2\)+1>0

b,=  -(-4x+x\(^2\))-5

=    -(4-4x+x\(^2\))-5+4

=     -(2-x)\(^2\)-1

Mà  -(2-x)\(^2\)\(\le\)0

nên -(2-x)\(^2\)-1<   0

Võ Hoàng Tiên: Cảm ơn pạn nhiều lắm =)) nek :3 Hí Hí :)  Thankssssss 

Bài 4:

\(\frac{n^4+4}{n^2-2n+2}\)

\(=\frac{n^4+4+4n^2-4n^2}{n^2-2n+2}\)

\(=\frac{\left(n^2+2\right)^2-\left(2n\right)^2}{n^2-2n+2}=\frac{\left(n^2-2n+2\right)\left(n^2+2n+2\right)}{n^2-2n+2}=n^2+2n+2\)

Bài 3:

a: xét ΔBHA vuông tại H và ΔBAC vuông tại A có

\(\hat{HBA}\) chung

Do đó: ΔBHA~ΔBAC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=6^2+8^2=36+64=100=10^2\)

=>BC=10(cm)

ΔBHA~ΔBAC

=>\(\frac{BH}{BA}=\frac{BA}{BC}\)

=>\(BH\cdot BC=BA^2\)

=>BH=6^2/10=3,6(cm)

BH+HC=BC

=>HC=10-3,6=6,4(cm)

Câu 1:

a: 3x-6=0

=>3x=6

=>\(x=\frac63=2\)

b: 7+3x=9+x

=>3x-x=9-7

=>2x=2

=>\(x=\frac22=1\)

c: 8x+2<=x+23

=>8x-x<=23-2

=>7x<=21

=>x<=3

d:

ĐKXĐ: x<>0

\(\frac{x^2-6}{x}=x+\frac32\)

=>\(x-\frac{6}{x}=x+\frac32\)

=>\(-\frac{6}{x}=\frac32=\frac{-6}{-4}\)

=>x=-4(nhận)

\(x^7+x^5+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^5-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)

Thank you bạn nhiều nha!

Câu 27: B

Câu 28: B

Câu 29: A

Câu 30: D

Câu 31: A
Câu 32: B