Bài 1 :

a ) Ta có :

\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)

b ) Ta có :

\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

= 270 nha bạn

1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+8+9+9=2+4+6+8+10+12+14+16+18

                                                                        =(2+8)+(4+6)+10+(12+18)+(14+16)

                                                                        =10+10+10+30+30

                                                                        =30+30+30

                                                                        =30x3

                                                                        =90

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 \(N=\frac{7^{32}-1}{3}\)

\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)

\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)

\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)

Đề là gì vậy bạn?

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