Bài 1: a) x+x+x+x+x=5x tại x=- 5 ta được 5x=5*-5=-25

b) x-3+x-3+x-3+x-3=4(x-3) tại x=-5 ta được 4(x-3)=4(-5-3)=-32

a) (a-b)-(b-c)+(c-a)-(a-b-c)=a-b-b+c+c-a-a+b+c=-a-b+3c

b) (-a+b+c)-(a-b+c)-(-a+b-c)=-a+b+c-a+b-c+a-b+c=-a+b+c

c) x+12+(-18)+30=x+12-18+30=x+24

câu 1:a) \(5^x=125=5^3\Leftrightarrow x=3\)

b) \(3^{2x}=81=3^4\Leftrightarrow2x=4\Rightarrow x=\frac{4}{2}=2\)

c) \(5^{2x-3}-2\cdot5^2=5^2\cdot3\Leftrightarrow5^{2x-3}=5^2\cdot3+2\cdot5^2=125\)=\(5^3\)

=>2x-3=3 => x=(3+3)/2=3

a) khi m=1 thì ta có : 1x-y=3 (1)

2x+1y=9 (2)

(1)+(2) ta được 3x=3+9=12 =>x=12/3=4

thế x=4 vao (1) ta được 4-y=3=> y=4-3=1

Vậy x=4 ; y=1

b) 3(x-8)-2(x+1)=20

<=>3x-24-2x-2-20=0

<=>x-46=0 => x=46

Vậy x=46

a) x-34=(76-x)+108

<=>x-34-76+x-108=0

<=>2x-218=0=>x=218/2=109 .vậy x=109

ta có \(abc=n^2-1=\left(n-2\right)^2\)

\(\Leftrightarrow n^2-1-\left(n-2\right)^2=0\)

\(\Leftrightarrow n^2-1-n^2+4n-4=0\)

\(\Leftrightarrow4n-5=0\Rightarrow n=\frac{5}{4}\)

vậy abc=\(n^2-1=\left(\frac{5}{4}\right)^2-1=\frac{9}{16}\)

A=\(1+\frac{1}{2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}+\frac{1}{100}\)

=\(\left(1+\frac{1}{2}+\frac{1}{100}\right)+\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\right)\)

=\(\frac{151}{100}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

=\(\frac{151}{100}+\frac{1}{3}-\frac{1}{100}\)

=\(\frac{11}{6}\)

Thực hiện phép chia bình thường ta được dư bằng 6

Để A nguyên thì \(\left(x^2+2\right)⋮\left(x+2\right)\Leftrightarrow6⋮\left(x+2\right)\)

\(\Leftrightarrow\left(x+2\right)\inƯ\left(6\right)=\left\{|^+_-1;|^+_-2;|^+_-3;|^+_-6\right\}\)

lâp bảng :

Vậy x={-8;-5;-4;-3;-1;0;1;4}

a)y²-25y+24

=(y²-24y)-(y-24)

=y(y-24)-(y-24)

=(y-24)(y-1)

b)7y²-3y-10

=(7y²-10y)+(7y-10)

=y(7y-10)+(7y-10)

=(7y-10)(y+1)

c)5y²-15y+10

=5(y²-3y+2)

=5[(y²-2y)-(y-2)]

=5[y(y-2)-(y-2)]

=5(y-2)(y-1)