\(\left(x+1\right)^2+\left(2x-1\right)\left(2x+1\right)-5\left(x-1\right)=77\)

\(\Rightarrow x^2+2x+1+\left(2x\right)^2-1^2-5x+5=77\)

\(\Rightarrow x^2+2x+1+4x^2-1-5x+5=77\)

\(\Rightarrow5x^2-3x+5-77=0\)

\(\Rightarrow5x^2-3x-72=0\)

.....

Ta có : \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

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

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

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

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^{32}-1\right)-2^{32}\)

\(=-1\)

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32

=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32=(2^4-1)(2^4+1)(2^8+1)(2^16+1)-2^32

=(2^8-1)(2^8+1)(2^16+1)-2^32=(2^16-1)(2^16+1)-2^32=2^32-1-2^32=-1

\(x\) có 2 trường hợp:

TH1:

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

TH2:

\(x=0\)

a,Để phép chia thực hiện đc<=>x^n<=x^5=>n<=5=>n=(0;1;2;3;4;5)

                                                    y^n<=y=>n<=1=>n=(1;0)

Từ hai ý trên=>n=(0;1)

b,,Để phép chia thực hiện đc<=>x^n+1<=x^2=>n+1<=2=>n=(0;1)

                                                    y^n=1<=y^2=>n+1<=2=>n=(0;1)

Từ hai ý trên =>n=(0;1)

đk: \(x\ge-1\)

-xét x bằng 0 (tm)

-xét x khác 0=>phương trình có nghiệm khi x<0,khi đó ta có:

\(x+2.\sqrt{2.x^2.\left(x+1\right)}=0\) mà x < 0 nên khi rút gọn cho x ta có:

\(1-2.\sqrt{2\left(x+1\right)}=0\) => giải ra ta có  x=\(\frac{-7}{8}\) (tm).     vậy phương trình có 2 nghiệm là 0 và\(\frac{-7}{8}\)

1+2=3

tôi nè

=3

mk nè

kb nha

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