2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

a: \(A=2\cdot2^2-\dfrac{1}{3}\cdot9=8-3=5\)

b: \(B=\dfrac{1}{2}a^2-3b^2=\dfrac{1}{2}\cdot4-3\cdot\dfrac{1}{9}=2-\dfrac{1}{3}=\dfrac{5}{3}\)

Thay x = 2 và y=9

A = 2.2² -\(\dfrac{1}{3}\).9

=  2.4 -\(\dfrac{1}{3}.9\)

= 8 - 3

= 5

Thay a = -2 và b = \(-\dfrac{1}{3}\)

B = \(\dfrac{1}{2}.\left(-2\right)^2-3.\left(\dfrac{-1}{3}\right)^2\)

B = \(\dfrac{1}{2}.4-3.\dfrac{1}{9}\)

B = \(2-\dfrac{1}{3}\)

B = \(\dfrac{5}{3}\)

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

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=x^5-32

Khi x=3 thì A=3^5-32=243-32=211

b: B=x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+x^7-x^6+x^5-x^4+x^3-x^2+x-1

=x^8-1

=2^8-1=255

gúp mình với

A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

a, Với x = 3 và y = -2 ta có:

\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|3\right|\right)+\left(-2\right)\)

\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-3\right)-2\)

\(A=\dfrac{3}{2}+\dfrac{4}{9}.3-2\)

\(A=\dfrac{3}{2}+\dfrac{4}{3}-2\)

\(A=\dfrac{5}{6}\)

 Với x = 3 và y = -3 ta có:
\(B=\left|2.3-1\right|+\left|3.\left(-3\right)+2\right|\)

\(B=\left|5\right|+\left|-7\right|\)

\(B=5+7=12\)

Hoctot ! ko hiểu chỗ nào cứ hỏi cj nhé

E cảm ơn cj

\(=1-\dfrac{17^{12}\cdot2^6\cdot3^2\cdot7}{17^6\cdot\left(3^4\cdot5^2-2^2\cdot1\right)}\)

\(=1-\dfrac{2^6\cdot3^2\cdot7}{81\cdot25-4\cdot1}\)

\(=1-\dfrac{4032}{2021}=-\dfrac{2011}{2021}\)

a: A=5x^2y-5x^2y-3xy+2xy+xy+x^4y^2+1+x^2

=x^4y^2+x^2+1

Khi x=-1 và y=1 thì A=(-1)^4*1^2+(-1)^2+1=3

b: A=x^2(x^2y^2+1)+1>=1>0 với mọi x,y

=>A luôn dương với mọi x,y