Có \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(x-\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)

\(\Leftrightarrow\left[x^2-\left(\sqrt{x^2+1}\right)^2\right]\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)

\(\Leftrightarrow-y-\sqrt{y^2+1}=x-\sqrt{x^2+1}\) (1)

Lại có:\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)\left(y-\sqrt{y^2+1}\right)=y-\sqrt{y^2+1}\)

\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left[y^2-\left(\sqrt{y^2+1}\right)^2\right]=y-\sqrt{y^2+1}\)

\(\Leftrightarrow-x-\sqrt{x^2+1}=y-\sqrt{y^2+1}\)  (2)

Từ (1) và (2) cộng vế với vế có:

\(-\left(y+x\right)-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)=x+y-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)\)

\(\Leftrightarrow2\left(x+y\right)=0\)

\(\Leftrightarrow x+y=0\) hay S=0

Vậy...

\(S=\left(1+\dfrac{2a}{3b}\right)\left(1+\dfrac{2b}{3c}\right)\left(1+\dfrac{2c}{3d}\right)\left(1+\dfrac{2d}{3a}\right)\)

có \(1+\dfrac{2a}{3b}\ge2\sqrt{\dfrac{2a}{3b}}\)(BDT AM-GM)

\(=>1+\dfrac{2b}{3c}\ge2\sqrt{\dfrac{2b}{3c}}\)

\(=>1+\dfrac{2c}{3d}\ge2\sqrt{\dfrac{2c}{3d}}\)

\(=>1+\dfrac{2d}{3a}\ge2\sqrt{\dfrac{2d}{3a}}\)

\(=>S\ge16\sqrt{\dfrac{2a.2b.2c.2d}{3a.3b.3c.3d}}=16\sqrt{\dfrac{16abcd}{81abcd}}=16\sqrt{\dfrac{16}{81}}=\dfrac{64}{9}\)

rảnh mà sao ko dngf phân số

\(tacó:...\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}>\frac{1}{3.2}=\frac{1}{\left(1+2.1\right).2.1}\) 

\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}>\frac{1}{5.4}=\frac{1}{\left(1+2.2\right).2.2}\) 

\(\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}>\frac{1}{7.6}=\frac{1}{\left(1+2..3\right).2.3}\) 

....

\(\frac{1}{49.\left(\sqrt{48}+\sqrt{49}\right)}>\frac{1}{49.48}=\frac{1}{\left(1+2.48\right).2.48}\) 

cộng vế theo vế ta đươc S =\(\frac{1}{\left(1+2.1\right).2}+\frac{1}{\left(1+2.2\right).2.2}+...+\frac{1}{\left(1+2.48\right).48.2}\)

\(=\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{10}+\frac{1}{21}+\frac{1}{36}+...+\frac{1}{4656}\right)\)  <  \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4656}\right)\)

mà lại có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+..+\frac{1}{4656}\) 

=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9312}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{96.97}\) 

             = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{97}=\frac{1}{2}-\frac{1}{97}=\frac{95}{194}\)  

vậy S < \(\frac{95}{194}\) 

mà \(\frac{95}{194}< \frac{3}{7}\) 

=> S < \(\frac{3}{7}\)

KẾT LUẬN  : S <\(\frac{3}{7}\)

Đặt \(Q\left(x\right)=P\left(x\right)-3x-2\)

\(\Rightarrow Q\left(1\right)=Q\left(2\right)=Q\left(4\right)=0\)

\(\Rightarrow Q\left(x\right)\) có 3 nghiệm \(x=\left\{1;2;4\right\}\)

Do \(P\left(x\right)\) bậc 4 và có hệ số cao nhất bằng 1 \(\Rightarrow Q\left(x\right)\) cũng là đa thức bậc 4 có hệ số cao nhất bằng 1

\(\Rightarrow Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-x_0\right)\) với \(x_0\in R\)

\(\Rightarrow P\left(x\right)=Q\left(x\right)+3x+2=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-x_0\right)+3x+2\)

\(\Rightarrow P\left(5\right)=12\left(5-x_0\right)+17\) ; \(P\left(-1\right)=-30\left(-1-x_0\right)-1\)

\(\Rightarrow S=60\left(5-x_0\right)+85-60\left(-1-x_0\right)-2=443\)

Chọn B

B nhá bạn 

a)

\(f\left(-2\right)=\left(-2\right).\left(-2\right)+3=4+3=7\)

\(f\left(-1\right)=\left(-2\right).\left(-1\right)+3=2+3=5\)

\(f\left(0\right)=\left(-2\right).0+3=0+3=3\)

\(f\left(-\frac{1}{2}\right)=\left(-2\right).\left(-\frac{1}{2}\right)+3=1+3=4\)

\(f\left(\frac{1}{2}\right)=\left(-2\right).\frac{1}{2}+3=\left(-1\right)+3=2\)

Câu b thì bạn cứ thế số vào và làm tương tự vậy.

chúc bạn học tốt

b)g(-1)=(-1)²-1=1-1=0

g(0)=0²-1=0-1=-1

g(1)=1²-1=1-1=0

g(2)=2²-1=4-1=3