\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{\sqrt{6}+1}=1\)

\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{\sqrt{6}+1}=\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\frac{5\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}\)

\(=\sqrt{6}-\frac{5\left(\sqrt{6}-1\right)}{6-1}=\sqrt{6}-\sqrt{6}+1=1\)

\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{8\left(\sqrt{5}-1\right)}{\left(1+\sqrt{5}\right)\left(\sqrt{5}-1\right)}+3\sqrt{5}\)

\(=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(\sqrt{5}-1\right)}{5-1}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)

Trả lời :

= 5

Nhs

Hok T~

Ta có: \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)

\(\left(y-z\right)\ge0\Leftrightarrow y^2+z^2\ge2yz\)

\(\left(z-x\right)^2\ge0\Leftrightarrow z^2+x^2\ge2zx\)

\(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\)

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2+1\ge2y\)

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2+1\ge2z\)

Cộng lại vế với vế ta được: 

\(3\left(x^2+y^2+z^2\right)+3\ge2xy+2yz+2zx+2x+2y+2z\)

\(\Leftrightarrow Q\ge\frac{2\left(x+y+yz+xy+yz+zx\right)-3}{3}=3\)

Dấu \(=\)khi \(x=y=z=1\).

Ta có: \(x+y+z+xy+yz+xz\le x+y+z+\frac{\left(x+y+z\right)^2}{3}\)

=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)\ge3.6=18\)

<=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)

<=> \(\left(x+y+z-3\right)\left(x+y+z+6\right)\ge0\)

<=> \(x+y+z\ge3\)(vì x + y + z + 6 > 0 vì x,y,z > 0)

Do đó: \(Q=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

Dấu "=" xảy ra<=> x  = y= z và x + y + z = 3 <=> x = y = z = 1

Vậy MinQ = 3 <=> x = y= z = 1

23442432

Bạn tham khảo bài này nhé !

 \(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)

Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)

=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)

 \(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

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

Do đó: N = (-1)²⁰¹⁹ + 3.(-1)²⁰²⁰ - 2.(-1)²⁰²¹ = -1 + 3 + 2 = 4

Ta có: \(VT=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\)

=> \(VT^2=\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

\(VT^2=2\sqrt{5}+2\sqrt{5-4}=2\sqrt{5}+2=2\left(\sqrt{5}+1\right)\)

=> \(VT=\sqrt{\sqrt{5}+1}.\sqrt{2}=VP\) => VT = VP (đpcm)

\(T=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{9-12\sqrt{5}+20}}}\)

\(T=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

\(T=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

\(\Leftrightarrow\orbr{\begin{cases}3x-4y=2\\4x-3y=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}12x-16y=8\\12x-9y=-24\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-7y=32\\12x-9y=-24\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=-\frac{32}{7}\\12x-\frac{288}{7}=-24\end{cases}}\)

Tự giải tiếp nha