a) \(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)

\(\Leftrightarrow x-\sqrt{3}=\pm\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{3}=-\frac{\sqrt{3}}{2}\\x-\sqrt{3}=\frac{\sqrt{3}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{2}\\\frac{3\sqrt{3}}{2}\end{cases}}\)

Nghiệm cuối cùng là : \(x_1=\frac{\sqrt{3}}{2};x_2=\frac{3\sqrt{3}}{2}\)

b) || 6x - 2  | - 5 | = 2016. x -2017 

<=> || 6x - 2 | -5 | -2016x = -2017

<=> \(\orbr{\begin{cases}\left|6x-2\right|-5-2016.x=-2017,\left|6x-2\right|-5\ge0\\-\left(\left|6x-2\right|-5\right)-2016x=-2017,\left|6x-2\right|-5< 0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1,x\in\left[-\infty,-\frac{1}{2}\right];\left[\frac{7}{6};+\infty\right]\\x=\frac{1012}{1011},x\in\left[-\frac{1}{2},\frac{7}{6}\right]\end{cases}}\)

<=>\(\orbr{\begin{cases}x\in\varnothing\\x=\frac{1012}{1011}\end{cases}}\)

Vậy x = \(\frac{1012}{1011}\)

a, Ta có:
\(\orbr{\begin{cases}x-\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{4}}\\x-\sqrt{\frac{3}{4}}=-\sqrt{\frac{3}{4}}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\sqrt{\frac{3}{4}}\\x=0\end{cases}}}\)

mình xin lỗi , mình ghi sai đề

a)\(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)

Ta có :   \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)

    \(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)

        nhân theo vế của ( 1 ) ; ( 2 ) , ta có :

     \(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)

    \(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)

  rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :

     \(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)

     \(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)

     \(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\) 

       \(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)

  A = 2017 

 ( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :)   )

2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)

\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)

\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)

\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)

\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)

Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)

\(\Leftrightarrow x=2015;y=2016;z=2017\)

3/ \(\sqrt{\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}}\)

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

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

\(=|\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}|\) là số hữu tỉ

Bài 2:

a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)

Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)

\(\Rightarrow4x+12=6x\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\)

Vậy x = 6

b) Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)

\(=\frac{14-5}{8}=\frac{9}{8}\)

+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)

+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)

+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)

Vậy ...

c) \(5^x+5^{x+1}+5^{x+2}=3875\)

\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)

\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)

\(\Rightarrow5^x.31=3875\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

Vậy x = 3

Bài 1 hình như sai đề bạn ạ?

\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)

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

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

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

\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)

\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)

\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)

\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)

\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)

\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)

Đặt \(\hept{\begin{cases}\sqrt{2+x}=a\\\sqrt{2-x}=b\end{cases}\Rightarrow}a^2+b^2=4\)

\(\Rightarrow C=\frac{\sqrt{2ab}.\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2ab}.\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}\)

\(=\sqrt{2ab}.\left(a-b\right)=\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{2+x}-\sqrt{2-x}\right)\)

b)

)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

= \(\frac{2}{2-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\)

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

=\(\frac{4+2\sqrt{5}-4+2\sqrt{5}}{2^2-\sqrt{5}^2}\)

=\(\frac{4\sqrt{5}}{4-5}\)

=\(\frac{4\sqrt{5}}{-1}\)

\(-4\sqrt{5}\)