a, A = (-125).(-13).(-a) ; a=8

=> A = (-125).(-13).(-8)

=> A = [(-125).(-8)].(-13)

=> A = 1000.(-13)

=> A = -13000

vậy A = -13000 với a = 8

b,E = (-1).(-2).(-3).(-4).(-5).b ;  b = 20

=> E =  (-1).(-2).(-3).(-4).(-5).20

=> E = [(-1).(-2).(-3).(-4)]. [-5.20]

=> E = 24.(-100)

=> E = -2400

vậy E = -2400 với b = 20

a)-13000;b)-2400

A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)

\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)

\(=\dfrac{7}{20}-\dfrac{9}{40}\)

\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)

B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)

\(=\dfrac{5}{6}+\dfrac{11}{36}\).

\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)

C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)

\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)

a: =7/20-5/8+2/5

=14/40-25/40+16/40

=5/40=1/8

b: =5/6+5/9-1/4

=30/36+20/36-9/36

=41/36

c: =9/10-2/5+3/10+7/20

=12/10-2/5+7/20

=7/20+6/5-2/5

=7/20+4/5

=7/20+16/20

=23/20

a, a x 6 = 3 x 6 = 18

b, a + b = 4 + 2 = 6

c, b + a = 2 + 4 = 6

d, a - b = 8 - 5 = 3

e, m x n = 5 x 9 = 45

a. 18

b. 6

c. 6

d. 3

e. 45

Bài 1 : 

\(N=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....

a)A= 20-(x+2)-4x+25

Thay số

A= 20-(-2+2)-4.-2+25

 =  20-0-4.-2+25

 = 20-0-(-8)+25

=20-(-8)+25

=28+25

= 53

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

\(\Leftrightarrow x-4=\sqrt{x}+2\)

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9