Tìm nghiệm nguyên dương của phương trình: x^2+(x+y)^2=(x+9)^2 - Đại số - Diễn đàn Toán học

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{28-10\sqrt{3}}}}\)

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

\(=\sqrt{5\sqrt{3}+\sqrt{5\left(5-\sqrt{3}\right)}}=\sqrt{5\sqrt{3}+\sqrt{25-5\sqrt{3}}}\)

Trần Đức Thắng lm nốt đi

Ta có \(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}=1+\dfrac{a+b}{ab}+\dfrac{1}{ab}=1+\dfrac{a+b+1}{ab}=1+\dfrac{1+1}{ab}=1+\dfrac{2}{ab}\)

Áp dụng bđt cosi ta có

\(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\Leftrightarrow ab\le\dfrac{1}{4}\Leftrightarrow\dfrac{2}{ab}\ge8\Leftrightarrow1+\dfrac{2}{ab}\ge9\Leftrightarrow A\ge9\)

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(a=b=0,5\)

Vậy GTNN của A là 9 và xảy ra khi a=b=0,5

\(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)

\(A=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}\)

\(A=1+\dfrac{a+b}{ab}+\dfrac{1}{ab}\)

Mà a+b=1

nên \(A=1+\dfrac{1}{ab}+\dfrac{1}{ab}=1+\dfrac{2}{ab}\)

Ta có:

a+b=1

Áp dụng bđt Cosi

\(a+b\ge2\sqrt{ab}\Rightarrow1\ge2\sqrt{ab}\)

\(\Rightarrow1\ge4ab\Leftrightarrow ab\le\dfrac{1}{4}\)

Ta có:

\(A=1+\dfrac{2}{ab}\ge1+\dfrac{\dfrac{2}{1}}{4}=1+8=9\)

Dấu bằng xảy ra khi \(\) \(\left\{{}\begin{matrix}a+b=1\\a=b\end{matrix}\right.\)

\(\Rightarrow a=b=\dfrac{1}{2}\)

ab+bc+ca=3ac hay ab+bc+ca=3abc

Cứ phải cảnh giác bạn à:

không biết hay vô tình hay hưu ý nữa nhưng các câu hỏi sai xuất hiện rất nhiều

khi hỏi lại, không thấy phải hồi. hay là người hỏi cũng chưa hiểu câu hỏi

C3 

Đặt \(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\)

\(N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\)

Ta có : \(M+N=\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)+\left(\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\right)\)

\(=\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{c}{c+a}+\frac{a}{c+a}\right)+\left(\frac{a}{a+b}+\frac{b}{a+b}\right)\)

\(=\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a+b}{a+b}=1+1+1=3\)

Ta có :\(+)M+S=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\frac{b+a}{b+c}+\frac{c+b}{c+a}+\frac{a+c}{b+a}\)

Hoàn toàn tương tự :\(+)N+S=\frac{a+c}{b+c}+\frac{b+a}{c+a}+\frac{b+c}{b+a}\)

Áp dụng Bất đẳng thức Cauchy cho 2 số không âm ta được :

\(\frac{b+a}{b+c}+\frac{c+b}{c+a}+\frac{a+c}{b+a}\ge3\sqrt[3]{\frac{\left(b+a\right)\left(c+b\right)\left(a+c\right)}{\left(b+c\right)\left(c+a\right)\left(b+a\right)}}=3\)

\(\frac{a+c}{b+c}+\frac{b+a}{c+a}+\frac{b+c}{b+a}\ge3\sqrt[3]{\frac{\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(b+c\right)\left(c+a\right)\left(b+a\right)}}=3\)

Cộng theo vế các bất đẳng thức cùng chiều ta được :

\(M+N+2S\ge3+3=6\)

\(< =>3+2S\ge6< =>2S\ge6-3=3< =>S\ge\frac{2}{3}\)

Vậy ta có điều phải chứng minh

\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

\(\ge\left(a+b+c\right)\frac{9}{b+c+a+c+a+b}-3\)

\(=\frac{9}{2}-3=\frac{3}{2}\)

Dấu "=" xảy ra <=> a = b = c

Nguyễn Ngọc Quý sai ròi :

a) \(\sqrt{\left(2+\sqrt{5}\right)^2}=l2+\sqrt{5}l=2+\sqrt{5}\)

\(\sqrt{\left(3-\sqrt{15}\right)^2}=l3-\sqrt{15}l=\sqrt{15}-3\)

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

\(\sqrt{\left(3-\sqrt{15}\right)^2}=3-\sqrt{15}\)

\(B=\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}}-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)

\(\Rightarrow B=\dfrac{\sqrt{1}}{\sqrt{3}}+\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)

\(\Rightarrow B=\left(\dfrac{1}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}\right)+\dfrac{3-2\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\sqrt{3}\)

\(\Rightarrow B=\dfrac{3}{\sqrt{3}}+\dfrac{4-2\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\sqrt{3}\)

\(\Rightarrow B=\sqrt{3}+\dfrac{4-2\sqrt{3}}{\sqrt{3^2}-1^2}-\sqrt{3}\)

\(\Rightarrow B=\dfrac{4-2\sqrt{3}}{2}\)

\(\Rightarrow B=2-\sqrt{3}\)

⇔ \(B=\dfrac{1}{\sqrt{3}}+\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{2}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)

⇔ \(B=\dfrac{\text{3}}{\sqrt{3}}+\dfrac{3-2\sqrt{3}+1}{3-1}-\sqrt{3}\)

⇔ \(B=\sqrt{3}+\dfrac{\text{4}-2\sqrt{3}}{\text{2}}-\sqrt{3}\)

⇔ \(B=\dfrac{2\left(2-\sqrt{3}\right)}{2}=2-\sqrt{3}\)

Ta có:

VP= 3-2√3 + 1(vì 3 + 1 =4)

     =√3^2 - 2√3 + √1^2

     = (√3 - √1)^2 

a: ĐKXĐ: \(x\le0\)

b: ĐKXĐ: \(x\le2\)

\(a,ĐK:-3x\ge0\Leftrightarrow x\le0\left(-3< 0\right)\\ b,ĐK:4-2x\ge0\Leftrightarrow-2x\ge-4\Leftrightarrow x\le2\\ c,ĐK:\dfrac{1}{2x-5}\ge0\Leftrightarrow2x-5>0\left(1>0;2x-5\ne0\right)\\ \Leftrightarrow x>\dfrac{5}{2}\\ d,ĐK:\dfrac{4x+7}{-3}\ge0\Leftrightarrow4x+7\le0\left(-3< 0\right)\Leftrightarrow x\le-\dfrac{7}{4}\)