a) ( m + 7 ) 2 ( m + 9 ) 2           b) ( n + 7 ) 2 ( n + 8 ) 2

\(a,ĐK:x\ne\pm1\\ b,B=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\\ c,B=-\dfrac{1}{2}\Leftrightarrow2\left(x+1\right)=-2\Leftrightarrow x+1=-1\Leftrightarrow x=-2\left(tm\right)\)

đây là phân số hay sao bn

1) ta có \(\left(x+y\right)^2=x^2+2xy+y^2.\)

                                \(=\left(x^2+y^2\right)+2xy\)

                                \(=20+2.8\)(theo giả thiết x^2+y^2=20 , xy=8)

                                \(=36\)

Vậy với x^2+y^2=20, xy=8 thì (x+y)^2=36

2) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Rightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

  \(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Leftrightarrow3M=\left[\left(2^2\right)^2-1^2\right]\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Leftrightarrow3M=\left[\left(2^4\right)^2-1^2\right]\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left[\left(2^8\right)^2-1^2\right]\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^{16}\right)^2-1^2\)

\(\Leftrightarrow3M=2^{32}-1\)

\(\Rightarrow M=\frac{2^{32}-1}{3}\)

RÚT GỌN BIỂU THỨC N BẠN LÀM TƯƠNG TỰ NHA 

\(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

 \(\Rightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(...\)

\(...\)

Kết quả rút gọn \(N=\frac{7^{32}-1}{3}\)

a: \(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}=\dfrac{x-1}{x+1}\)

b: \(=\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\)

c: \(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)

Bạn lưu ý viết đề bằng công thức toán để được hỗ trợ tốt hơn. Viết thế này nhìn khá khó đọc.

Để viết công thức toán bạn nhấn biểu tượng $\sum$ góc trái khung soạn thảo.

cíuuu

a: =căn 3+căn 5-căn 3=căn5
b: \(=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

\(=\left|\sqrt{x-2}-1\right|\)

Nguyên Đinh Huynh Ronaldo lúc nào cũng dễ

3x(x – 2) – 5x(1 – x) – 8( x 2  – 3)

= 3x.x + 3x .( -2) – [5x.1 + 5x. (- x)] – [8 x 2  + 8.(- 3)]

= (3 x 2  – 6x) – (5x – 5 x 2 ) – (8 x 2  – 24)

= 3 x 2  – 6x – 5x + 5 x 2  – 8 x 2 + 24

= ( 3 x 2  +5 x 2  – 8 x 2 )- ( 6x + 5x) + 24

= - 11x + 24