cua lop may

trời tui mới lớp 6

\(\sqrt{3x+4}=\sqrt[3]{y^3+5y^2+7y+4}\)(\(x\ge-1\),VP>0)

=>\(3x+4=\sqrt[3]{\left(y^3+5y^2+7y+4\right)^2}\)

Do 3x+4 là số nguyên nên

 \(\sqrt[3]{\left(y^3+5y^2+7y+4\right)^2}\in Z\)=>\(\sqrt[3]{y^3+5y^2+7y+4}\in Z\)(1)

Ta có \(2y^2+4y+3=2\left(y+1\right)^2+1>0\)

=> \(y^3+5y^2+7y+4>y^3+3y^2+3y+1=\left(y+1\right)^3\)

=> \(y+1< \sqrt[3]{y^3+5y^2+7y+4}\)

Làm tương tự ta chứng minh được \(\sqrt[3]{y^3+5y^2+7y+4}< y+5\)

=> \(y+1< \sqrt[3]{y^3+5y^2+7y+4}< y+5\)

Kết hợp với (1)=> \(\orbr{\begin{cases}\sqrt[3]{y^3+5y^2+7y+4}=y+2\\\sqrt[3]{y^3+5y^2+7y+4}=y+3hoacy+4\end{cases}}\)

=> \(y\in\left\{-3;-1\right\}\)

+ y=-3 => x=-1

+y=-1 => x=-1

Vậy nghiệm nguyên của phương trình là \(\left(x;y\right)=\left(-1;-3\right),\left(-1;-1\right)\)

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

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

\(A=\sqrt{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}.\sqrt{\sqrt{3}+1}.\sqrt{2}\)

\(A=\sqrt{3-1}.\sqrt{\sqrt{3}+1}.\sqrt{2}\)

\(A=\sqrt{2}.\sqrt{2}.\sqrt{\sqrt{3}+1}=2.\sqrt{\sqrt{3}+1}\)

Vậy \(A=2\sqrt{\sqrt{3}+1}\).

\(a.\sqrt{50}+3\sqrt{72}-4\sqrt{128}+2\sqrt{162}=5\sqrt{2}+3\times6\sqrt{2}-4\times8\sqrt{2}+2\times9\sqrt{2}\)

\(=\left(5+18-32+18\right)\sqrt{2}=9\sqrt{2}\)

\(b.\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{16-8\sqrt{5}+5}=\sqrt{5}-2+\sqrt{\left(4-\sqrt{5}\right)^2}=\sqrt{5}-2+4+\sqrt{5}=2\)

\(c.\frac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-3\sqrt{4\times\frac{5}{2}}+\frac{12}{4-\sqrt{10}}=\sqrt{10}-3\sqrt{10}+\frac{12\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}\)

\(=-2\sqrt{10}+2\left(4+\sqrt{10}\right)=8\)