Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(D=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\)
Đặt a^2+6a=x
=>\(D=\sqrt{x\left(x+5\right)\left(x+8\right)+36}\)
\(=\sqrt{x\left(x^2+13x+40\right)+36}\)
\(=\sqrt{x^3+13x^2+40x+36}\)
=>\(D=\sqrt{x^3+9x^2+4x^2+36x+4x+36}\)
\(=\sqrt{\left(x+9\right)\left(x^2+4x+4\right)}\)
\(=\sqrt{\left(a^2+6a+9\right)\left(x+2\right)^2}\)
=|a+3|*|x+2| là số nguyên
Nhìn cái D cồng kềnh thế thôi chứ key vô cùng EZ.
\(D=\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)
\(=\sqrt{\left[a\left(a+6\right)\right]\left[\left(a+1\right)\left(a+5\right)\right]\left[\left(a+2\right)\left(a+4\right)\right]+36}\)
\(=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\)
Đặt \(a^2+6a=x\)
Ta có:
\(D=\sqrt{x\left(x+5\right)\left(x+8\right)+36}=\sqrt{x^3+13x^2+40x+36}\)
\(=\sqrt{\left(x+9\right)\left(x+2\right)^2}\)
Thay \(x=a^2+6a\) ta có:
\(D=\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a+6a+2\right)^2}=\left(a+3\right)\left(a+6a+2\right)\)
là số nguyên vs a nguyên khác 0 nha !
\(\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)
=\(\sqrt{\left(a\left(a+4\right)\left(a+5\right)\right).\left(\left(a+1\right)\left(a+2\right)\left(a+6\right)\right)+36}\)
\(\sqrt{\left(a^3+9a^2+20a\right).\left(a^3+9a^2+20a+12\right)+36}\)
Đặt a^3+9a^2+20a+6=k(k thuộc Z)
ta có\(\sqrt{\left(k-6\right)\left(k+6\right)+36}=\sqrt{k^2-36+36}=\sqrt{k^2}=k\)
Vì k thuộc Z
=>A thuộc Z
tick nha
Lời giải:
\(a(a+1)(a+2)(a+4)(a+5)(a+6)+36=[a(a+4)(a+5)][(a+1)(a+2)(a+6)]+36\)
\(=(a^3+9a^2+20a)(a^3+9a^2+20a+12)+36\)
\(=(a^3+9a^2+20a)^2+12(a^3+9a^2+20a)+36\)
\(=(a^3+9a^2+20a+6)^2\)
\(\Rightarrow \sqrt{a(a+1)(a+2)(a+4)(a+5)+36}=|a^3+9a^2+20a+6|\) có giá trị nguyên với mọi $a$ nguyên (đpcm)
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
\(a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36=\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36\)
Đặt \(a^2+6a=t\) ta có:\(t\left(t+5\right)\left(t+8\right)+36=t\left(t^2+13t+40\right)=t^3+13t^2+40t+36=\left(t+9\right)\left(t+2\right)^2\)
Do đó \(\sqrt{\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}=\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a^2+6a+2\right)^2}\)
\(=\left(a+3\right)\left(a^2+6a+2\right)\)(Dấu () ở đây là giá trị tuyệt đối nha)
Do đó với a nguyên thì \(\left(a+3\right)\left(a^2+6a+2\right)\)nguyên (Dấu () ở đây là giá trị tuyệt đối nha)
Vậy nếu a nguyên thì \(\sqrt{\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)nguyên