Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

\(D=\left(\dfrac{a}{c}\right)^{2016}+\left(\dfrac{c}{b}\right)^{2017}+\left(\dfrac{b}{a}\right)^{2018}=1+1+1=3\)

\(A=2x^2+9y^2-6xy+6x+2017\)

\(A=x^2+x^2+9y^2-6xy+6x+2017\)

\(A=\left(x^2-6xy+9y^2\right)+\left(x^2+6x+9\right)+2008\)

\(A=\left(x-3y\right)^2+\left(x+3\right)^2+2008\ge2008\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-3y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y\\x=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=-3\end{matrix}\right.\)

\(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3}{\sqrt{x}+3}-\dfrac{8}{\sqrt{x}+3}=1+\dfrac{8}{\sqrt{x}+3}\)

Đến đây thì đơn giản rồi nhé

\(\sqrt{16x+16}+\sqrt{9x+9}-\sqrt{25x+25}+2\sqrt{x+1}=8\)

\(\Rightarrow4\sqrt{x+1}+3\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=8\)

\(\Rightarrow\sqrt{x+1}\left(4+3-5+2\right)=8\)

\(\Rightarrow4\sqrt{x+1}=8\)

\(\Rightarrow\sqrt{x+1}=2\)

\(\Rightarrow x+1=4\)

\(\Rightarrow\)\(x=3\)

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

Chí công vô tư có ý nghĩa là?

A. Đem lại lợi ích cho tập thể.

B. Góp phần làm cho đất nước giàu mạnh, xã hội công bằng, dân chủ, văn minh.

C. Đem lại lợi ích cho cộng đồng xã hội.

\(A=\left|x+2\right|+\left|x+5\right|+\left|x-7\right|+\left|x-8\right|\)

\(A=\left|x+2\right|+\left|x+5\right|+\left|7-x\right|+\left|8-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(A\ge\left|x+2+x+5+7-x+8-x\right|=22\)

Dấu "=" xảy ra khi: \(\left(x+2\right)\left(x+5\right)\left(7-x\right)\left(8-x\right)\ge0\)

Nên \(\left\{{}\begin{matrix}x\ge-2\\x\ge-5\\x\le7\\x\le8\end{matrix}\right.\Leftrightarrow-2\le x\le7\)

\(\sqrt{9x+8}+\sqrt{4x-8}=20\)

\(\Rightarrow\left(\sqrt{9x+8}+\sqrt{4x-8}\right)^2=400\)

\(\Rightarrow9x+8+2\sqrt{\left(9x+8\right)\left(4x-8\right)}+4x-8=400\)

\(\Rightarrow13x+\sqrt{4\left(9x+8\right)\left(4x-8\right)}=400\)

\(\Rightarrow\sqrt{4\left(9x+8\right)\left(4x-8\right)}=400-13x\)

\(\Rightarrow4\left(9x+8\right)\left(4x-8\right)=\left(400-13x\right)^2\)

\(\Rightarrow\left(36x+32\right)\left(4x-8\right)=160000-10400x+169x^2\)

\(\Rightarrow36x\left(4x-8\right)+32\left(4x-8\right)=160000-10400x+169x^2\)

\(\Rightarrow144x^2-288x+128x-256=160000-10400x+169x^2\)

\(\Rightarrow144x^2-160x-256=160000-10400x+169x^2\)

\(\Rightarrow160000-10400x+169x^2-144x^2+160x+256=0\)

\(\Rightarrow\left(160000+256\right)+\left(169x^2-144x^2\right)+\left(160x-10400x\right)=0\)\(\Rightarrow160256+25x^2-10240x=0\)

\(\Rightarrow1048576-888320+25x^2-10240x=0\)

\(\Rightarrow1048576+25x^2-10240x=888320\)

\(\Rightarrow\left(1024-5x\right)^2=888320\)

\(\Rightarrow\left[{}\begin{matrix}1024-5x=\sqrt{888320}\\1024-5x=-\sqrt{888320}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1024-\sqrt{888320}}{5}\\x=\dfrac{1024+\sqrt{888320}}{5}\end{matrix}\right.\)