pt <=> \(\frac{3x-1}{x+2}=4\)

<=>  \(3x-1=4\left(x+2\right)\)

<=> \(3x-1=4x+8\)

<=> \(x=-9\)

VẬY \(x=-9\)

\(\sqrt{\frac{3x-1}{x+2}}=2\Leftrightarrow\frac{3x-1}{x+2}=4\)

\(\Leftrightarrow\frac{3x-1}{x+2}=\frac{4x+8}{x+2}\Leftrightarrow3x-1=4x+8\)

\(\Leftrightarrow-x=9\Leftrightarrow x=-9\)

Ap dung bdt Holder ta co

\(VP=\left(a^3+b^3+0^3\right)\left(b^3+y^3+0^3\right)\left(c^3+z^3+0^3\right)\ge\left(abc+xyz+0\right)^3=VT\)

P/s: Day la 1 he qua quen thuoc cua bdt Holder

Automa checking inequality. (Ảnh trong thống kê hỏi đáp)

Đặt: \(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

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

=> \(A^2=4-2\sqrt{4-3}\)

=> \(A^2=4-2\)

=> \(A^2=2\)

=> \(A=\sqrt{2}\)DO \(A>0\)

VẬY \(A=\sqrt{2}\)

mình có sửa lại đề 1 chút!

đặt \(T=\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

đặt \(u=a^4;v=b^6\)(a,b>0) ta có

\(T=\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}=\frac{a^4-8a^2b^2+4b^2}{a^2-2b^2+2ab}+3b^2\)

vậy \(T=\frac{a^4-8a^2b^2+4b^4}{a^2-2b^2+2ab}+3b^2=\frac{a^4-5a^2b^2-2b^4+6ab^3}{a^2-2b^2+2ab}=a^2-2ab+b^2\)

từ đó suy ra \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}\)

vì \(u^3\ge v^2\)nên \(\left|\sqrt[4]{u}-\sqrt[6]{v}\right|+\sqrt[6]{v}=\sqrt[4]{u}\)

\(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)

với u=1 ta có \(T=\sqrt{\frac{1-8\sqrt[6]{v^2}+4\sqrt[3]{v^2}}{1-2\sqrt[3]{v}+2\sqrt[6]{v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}\)

nếu \(1-2\sqrt[3]{v}+2\sqrt[6]{v}=0\)thì \(\sqrt[3]{v}=\frac{3+1}{2}>0\)

do \(v^2>1=u^3\), mâu thuẫn suy ra \(1-2\sqrt[3]{v}+2\sqrt[6]{v}\ne0\)

tóm lại với \(u^3\ge v^2\)và u,v\(\inℚ^+\)để \(\sqrt{\frac{u-8\sqrt[6]{u^3v^2}+4\sqrt[3]{v^2}}{\sqrt{u}-2\sqrt[3]{v}+2\sqrt[12]{u^3v^2}}+3\sqrt[3]{v}}+\sqrt[6]{v}=1\)cần và đủ là u=1 và v<1, v\(\inℚ^+\)được lấy tùy ý

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

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

\(=3-\sqrt{5}+2.\sqrt{3^2-5}+3+\sqrt{5}\)

\(=6+2.\sqrt{4}=6+2.2=6+4=10\)

hình như kết quả là 5 í!

Kết quả là 25

Bài làm:

Ta có: \(M=\sqrt{x^2+2x+5}=\sqrt{\left(x+1\right)^2+4}\)

Mà \(\left(x+1\right)^2+4\ge4\left(\forall x\right)\)

=> \(M\ge2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy \(M_{Min}=2\Leftrightarrow x=-1\)

\(M=\sqrt{x^2+2x+5}\)

\(\Leftrightarrow M=\sqrt{x^2+2x+1+4}\)

\(\Leftrightarrow M=\sqrt{\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Min M = 2 

\(\Leftrightarrow x=-1\)

Bài làm:

Ta có: \(\sqrt{3-2x}< 5\)

\(\Leftrightarrow\left|3-2x\right|< 25\)

\(\Leftrightarrow-5< 3-2x< 5\)

\(\Leftrightarrow3-\left(-5\right)>3-\left(3-2x\right)>3-5\)

\(\Leftrightarrow8>2x>-2\)

\(\Rightarrow-1< x< 4\)