`a)  (2^5 .3^2)/(2^3 .3)`

`= (2^2 .3)/1`

`= 4.3`

`=12`

`b)  (5^3 .6)/(5.2^2)`

`= (5^3 .2.3)/(5.2^2)`

`= (5^2 .3)/2`

`= 75/2`

`c)  (8.5^2-8.4^2)/(2^2 .3^2)`

`= (8.(5^2-4^2))/(4.9)`

`= (8.9)/(4.9)`

`= (2.4.9)/(4.9)`

`=2`

`d)  (7^4 .2^2+7^4 .3)/(49.26)`

`= (7^4.(2^2+3))/(7^2 .26)`

`= (7^4 .7)/(7^2 .26)`

`= (7^5)/(7^2. 26)`

`= (7^3)/26`

`= 343/26`

Ôi , bn sửa lại rồi à

nhưng bn vẫn bị lag

đây mà là toán lớp 5????????

Toán lớp 6 chứ

'Lớp 5 còn chưa học số mũ mà'

bn bị lag à

Ta có: \(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)

\(=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}\)

\(=-6\sqrt{3}\)

\(\Rightarrow A=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}=-4\sqrt{3}\)

⇔ A = \(3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\)

⇔ \(A=-6\sqrt{3}\)

Khi \(x>9\) thì:

\(A=x-9+7-x=-2\)

\(A=\left(x+2\right)^3-\left(x-2\right)^3-12x^2=x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2=16\)

\(A=\left(5a-5\right)^2+10\left(a-3\right)\left(1+a\right).3a\)

\(A=25a^2-50a+25+30a\left(a-3+a^2-3a\right)\)

\(A=25a^2-50a+25+30a^2-90a+30a^3-90a^2\)

\(A=30a^3-35a^2-140a+25\)

Ta có: \(A=\left(5a-5\right)^2+10\left(a-3\right)\left(a+1\right)\cdot3a\)

\(=25a^2-50a+25+30a\left(a^2-2a-3\right)\)

\(=25a^2-50a+25+30a^3-60a^2-90a\)

\(=30a^3-35a^2-140a+25\)

a) Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

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

\(=4x-3x^2+3\)

\(=-3x^2+4x+3\)

b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)

\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)

\(=5x^2-20-32x^2+48x-16+17\)

\(=-27x^2+48x-19\)

Bài 1: 

\(\dfrac{x^2-3}{x+\sqrt{3}}=\dfrac{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)

Bài 2: 

a) Ta có: \(A=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)

\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

b) Để A=16 thì \(\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

hay x=15

Viết latex cho dễ hiểu bn ơi

\(A=\dfrac{x^2-2x-3-x^2+x-1+4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{3x}{\left(x+1\right)\left(x^2+x+1\right)}\)