\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(2\left(a^2+b^2+c^2-ab-bc-ca\right)\)\(=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(0=4\left(a^2+b^2+c^2-ab-bc-ca\right)-\)\(2\left(a^2+b^2+c^2-ab-bc-ca\right)\)

<=>\(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\)

<=>a-b=b-c=c-a<=>a=b=c<=>a+b=2c

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)

a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)

\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)

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

b: \(x=\dfrac{2}{2-\sqrt{3}}=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)

Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)

c: C=-1

=>\(2\sqrt{x}-3=1\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

d: C>0

=>\(2\sqrt{x}-3< 0\)

=>\(\sqrt{x}< \dfrac{3}{2}\)

=>\(0< =x< \dfrac{9}{4}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)

@Nguyễn Việt Lâm

https://hoc24.vn/id/2782086

Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)

Thật vậy, BĐT trên tương đương:

\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)

\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)

\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)

Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)

\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)

\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Câu 2/ Gọi ước chung lớn nhất của a,c là q thì ta có:

a = qa₁; c = qc₁ (a₁, c₁ nguyên tố cùng nhau).

Thay vào điều kiện ta được:

 qa₁b = qc₁d

\(\Leftrightarrow\)a₁b = c₁d

\(\Rightarrow\)  d\(⋮\)a₁

\(\Rightarrow\)d = d₁a₁

Thế ngược lại ta được: b = d₁c₁

Từ đây ta có:

A = aⁿ + bⁿ + cⁿ + dⁿ = (qa₁)ⁿ + (qc₁)ⁿ + (d₁a₁)ⁿ + (d₁c₁)ⁿ

= (a​₁ ⁿ + c₁ ⁿ)(q ⁿ + d₁ ⁿ)

Vậy A là hợp số

\(D=\frac{4}{1^2}+\frac{4}{3^2}+....+\frac{4}{2015^2}\)

\(D=4+2.\left(\frac{2}{3.3}+\frac{2}{5.5}+....+\frac{2}{2015.2015}\right)\)

\(D< 4+2.\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{2013.2015}\right)\)

\(D< 4+2.\left(1-\frac{1}{2015}\right)\)

\(D< 6\)

mink chỉ làm được vậy thôi bạn ạ, sorry

\(D=\frac{2^2}{1^2}+\frac{2^2}{3^2}+....+\frac{2^2}{2015^2}\)

\(D=4+2\left(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2015^2}\right)\)

Ta có

\(\frac{2}{3^2}< \frac{2}{1.3}=1-\frac{1}{3}\)

\(\frac{2}{5^2}< \frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)

\(\frac{2}{7^2}< \frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)

  \(...........\)

\(\frac{2}{2015^2}< \frac{2}{2015.2017}=\frac{1}{2013}-\frac{1}{2015}\)

\(\Rightarrow\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+.....+\frac{2}{2015^2}< 1-\frac{1}{2015}< 1\)

\(\Rightarrow D< 4+2.1\)

\(\Rightarrow D< 6\)

Ta có:

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)

\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)

\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)

\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)

\(A=\frac{1.2018}{2017.2}\)

\(A=\frac{1009}{2017}\)

Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)

           \(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)

Vậy A>B

Tự cm bđt phụ: \(\frac{\left(m+n\right)^2}{x+y}\le\frac{m^2}{x}+\frac{n^2}{y}\)      Với x;y>0

Áp dụng ta có \(\frac{\left(b+c\right)^2}{b^2+c^2+a\left(b+c\right)}\le\frac{b^2}{b^2+ab}+\frac{c^2}{c^2+ab}=\frac{b}{a+b}+\frac{c}{a+c}\)

Tương tự có đpcm