\(\sqrt{x+7}-\sqrt{x-82}=x-2017\)

\(\Leftrightarrow\left(\sqrt{x+7}-45\right)-\left(\sqrt{x-82}-44\right)=x-2018\)

\(\Leftrightarrow\frac{x-2018}{\sqrt{x+7}+45}-\frac{x-2018}{\sqrt{x-82}+44}=x-2018\)

\(\Leftrightarrow\left(x-2018\right)\left(\frac{1}{\sqrt{x+7}+45}-\frac{1}{\sqrt{x-82}+44}-1\right)=0\)

\(\Rightarrow x=2018\)

X=2018 

Mình sẽ nghĩ cách giải sau

\(\sqrt{x+7}-45-\sqrt{x-82}+44=x-2018\)

\(\dfrac{\left(\sqrt{x+7}-45\right)\left(\sqrt{x+7}+45\right)}{\sqrt{x+7}+45}-\dfrac{\left(\sqrt{x-82}-44\right)\left(\sqrt{x-82}+44\right)}{\sqrt{x-82}+44}-\left(x-2018\right)=0\)

\(\dfrac{x-2018}{\sqrt{x+7}+45}-\dfrac{x-2018}{\sqrt{x-82}+44}-\left(x-2018\right)=0\) \(\left(x-2018\right)\left(\dfrac{1}{\sqrt{x+7}+45}-\dfrac{1}{\sqrt{x-82}+44}-1\right)=0\)

mà \(\dfrac{1}{\sqrt{x+7}+45}-\dfrac{1}{\sqrt{x-82}+44}-1< 0\)

\(\Rightarrow x=2018\)

\(x^4+\sqrt{x^2+2017}=2017\)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}=x^2+2017-\sqrt{x^2+2017}+\frac{1}{4}\)

\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2=\left(\sqrt{x^2+2017}-\frac{1}{2}\right)^2\)

\(\Leftrightarrow x^2+\frac{1}{2}=\sqrt{x^2+2017}-\frac{1}{2}\)(vì \(\sqrt{x^2+2017}>\frac{1}{2}\))

\(\Leftrightarrow x^2-\sqrt{x^2+2017}+1=0\)

\(\Leftrightarrow\left(x^2+2017-\sqrt{x^2+2017}+\frac{1}{4}\right)=\frac{8065}{4}\)

\(\Leftrightarrow\left(\sqrt{x^2+2017}-\frac{1}{2}\right)^2=\frac{8065}{4}\)

\(\Leftrightarrow\sqrt{x^2+2017}=\frac{\sqrt{8065}+1}{2}\)

\(\Leftrightarrow x^2=\frac{\left(\sqrt{8065}+1\right)^2}{4}-2017\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{\left(\sqrt{8065}+1\right)^2}{4}-2017}\\x=-\sqrt{\frac{\left(\sqrt{8065}+1\right)^2}{4}-2017}\end{cases}}\)

Cảm ơn bạn nha

từ a+b=3 => b=3-a

mặt khác: \(a^3-b^2=-3\)

=>\(a^3-\left(3-a\right)^2+3=0\)

\(\Rightarrow a^3-9+6a-a^2+3=0\)

\(\Rightarrow a^3-a^2+6a-6=0\)

\(\Rightarrow a^2\left(a-1\right)+6\left(a-1\right)=0\)

\(\Rightarrow\left(a^2+6\right)\left(a-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}a^2+6=0\\a-1=0\end{cases}\Rightarrow\hept{\begin{cases}a^2=-6\\a=1\end{cases}}}\)

=>a=1 vì \(a^2\ge0\)

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

\(\Rightarrow x-2=1\Rightarrow x=3\)

Vậy x=3

b) ta có: Đặt :\(\sqrt[3]{x-2}=a;\)    Đk: \(x\ge-1\)

                \(\sqrt{x+1}=b;b\ge0\)

ta có:\(\hept{\begin{cases}a+b=3\\a^3-b^2=-3\end{cases}}\)

đến đây dùng pp thế là đc rồi nhé!

thế như nào bạn mình hơi ngu

Đặt t=\(\sqrt{2019-x^{ }2}\)>0, nên \(t^2\)=2019-\(x^2\) hay \(x^2\)=2019-\(t^2\).

từ đề bài ta có: 2019-\(t^2\)-\(t^2\)-2017t=0

hay 2\(t^2\)+2017t-2019=0, nên t=1 và t=-2019/2<0 loại

t=1, nên \(x^2\)=2018, nên x=2018 hoặc x=-2018 thỏa điều kiện 2019-\(x^2\)>=0