\(\frac{7}{8}x-5\left(x-9\right)=\frac{20+1,5}{6}\)

2(x+3/5)=5-(13/5+x)

<=>2x+6/5=5-13/5-x

<=>2x+6/5-5+13/5+x=0

<=>3x-6/5=0

<=>3x=6/5

<=>x=2/3

Vậy phương trình có nghiệm duy nhất x=2/3

\(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)

\(\Leftrightarrow2x+\frac{6}{5}=5-\frac{13}{5}-x\)

\(\Leftrightarrow2x+\frac{6}{5}=\frac{12}{5}-x\)

\(\Leftrightarrow2x+\frac{6}{5}-\frac{12}{5}+x=0\)

\(\Leftrightarrow3x-\frac{6}{5}=0\)

\(\Leftrightarrow3x=\frac{6}{5}\)

\(\Leftrightarrow x=\frac{6}{15}=\frac{2}{5}\)

Chứng minh BĐT phụ :

\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

Thật vậy : \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) ( luôn đúng )

Áp dụng vào bài toán ta có : \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow2025\ge\left(x+y\right)^2\)

\(\Leftrightarrow-45\le x+y\le45\)

Vậy : \(min\left(x+y\right)=-45,max\left(x+y\right)=45\)

Bài này bạn chỉ cần chuyển vế biến đổi thôi là được , mình làm mẫu câu 2) :

\(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\)

\(\Leftrightarrow\frac{a^2n+b^2m}{mn}-\frac{\left(a+b\right)^2}{m+n}\ge0\)

\(\Leftrightarrow\frac{\left(m+n\right)\left(a^2n+b^2m\right)-\left(a^2+2ab+b^2\right).mn}{mn\left(m+n\right)}\ge0\)

\(\Leftrightarrow\frac{a^2mn+\left(bm\right)^2+\left(an\right)^2+b^2mn-a^2mn-2abmn-b^2mn}{mn\left(m+n\right)}\ge0\)

\(\Leftrightarrow\frac{\left(bm-an\right)^2}{mn\left(m+n\right)}\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow bm=an\)

Câu 3) áp dụng câu 2) để chứng minh dễ dàng hơn, ghép cặp 2 .

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

(ĐK a\(\ne-3\))

\(\Leftrightarrow40\left(a+3\right)-3\left(3a-1\right)-2\left(7a+2\right)=24\left(a+3\right)\)

\(\Leftrightarrow40a+120-9a+3-14a-4=24a+72\)

\(\Leftrightarrow7a=47\)

\(\Leftrightarrow a=\frac{47}{7}\)

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

\(\frac{10}{3}-\frac{3a-1}{4\left(a+3\right)}-\frac{7a+2}{6\left(a+3\right)}=2\)

\(40\left(a+3\right)-3\left(a-1\right)-2\left(7a+2\right)=24\left(a+3\right)\)

\(17a+119=24a+27\)

\(17a-24a=72-119\)

\(-7a=-47\)

\(a=\frac{47}{7}\)

C1 : Áp dụng BĐT Cô - si cho 3 số không âm ta được :

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

C2 : Sử dụng biến đổi tương đương :

Ta có :\(a^3+b^3+c^3\ge3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc\ge0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\) ( luôn đúng )

Do đó có : \(a^3+b^3+c^3\ge3abc\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Xét hiệu \(a^3+b^3+c^3-3abc\) ta có:

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)^3-3\left(a+b\right).c.\left(a+b+c\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right).c-3ab\right]\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab\right)\)

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

\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)

\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right]\)

\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)

Vì \(a,b,c\ge0\)\(\Rightarrow a+b+c\ge0\)

mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

\(\Rightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\ge0\)

hay \(a^3+b^3+c^3-3abc\ge0\)\(\Rightarrow a^3+b^3+c^3\ge3abc\)

Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=b=c=0\\a=b=c\end{cases}}\)\(\Leftrightarrow a=b=c\ge0\)