Question

Tính giá trị của biểu thức:

\(a,A=x^3+12x-8\)\(\text{ }\)với \(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)

\(b,B=x+y,\) biết \(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

\(c,C=\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2},\) biết \(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(d,D=x\sqrt{1+y^2}+y\sqrt{1+x^2},\) biết \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a\)

- @Toshiro Kiyoshi, @Akai Haruma, ....

Answer 1

a,Ta có :\(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)

\(\Rightarrow x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3\sqrt[3]{4\left(\sqrt{5}-1\right).4\left(\sqrt{5}+1\right)}.\left(\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\right)\)\(\Rightarrow x^3=8-3\sqrt[3]{16\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}.x\)

\(\Rightarrow x^3=8-3\sqrt[3]{64}.x\Rightarrow x^3=8-12x\)\(\Rightarrow x^3-12x+8=0\)

Vậy \(x^3+12x-8=0\)

Answer 2

b,\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)(1)

Ta có :\(3=\left(x^2+3\right)-x^2=\left(\sqrt{x^2+3}-x\right)\left(\sqrt{x^2+3}+x\right)\)(2)

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

Từ (1) và (2) ta suy ra :\(y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)

Từ (1) và (3) ta suy ra :\(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)

Cộng 2 đẳng thức trên vế theo vế ta được :

\(x+y+\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{x^2+3}+\sqrt{y^2+3}-x-y\)

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

Vậy B=0

Answer 3

c.Ta có :\(C=\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right).1\)

\(=\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)\(=\left(16-2x+x^2\right)-\left(9-2x+x^2\right)=7\)

Vậy C=7

Answer 4

d.Ta có \(D^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)

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

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

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

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

=\(a^2-1\)

Vậy D=\(a^2-1\)