\(2\sqrt{a}-a\sqrt{\dfrac{4}{a}}\)

\(=2\sqrt{a}-a.\dfrac{\sqrt{4}}{\sqrt{a}}\)

\(=2\sqrt{a}-a.\dfrac{2}{\sqrt{a}}\)

\(=2\sqrt{a}-2\sqrt{a}\)

\(=0\)

1) \(3x\left(x-4\right)-x+4=0\)

\(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

2) \(2x\left(2x+3\right)-2x-3=0\)

\(\Rightarrow2x\left(2x+3\right)-\left(2x+3\right)=0\)

\(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(3x\left(x-4\right)-x+4=0\\ \Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ 2x\left(2x+3\right)-2x-3=0\\ \Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(pt\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\\ \Leftrightarrow6x+20=0\Leftrightarrow x=-\dfrac{20}{6}=\dfrac{-10}{3}\)

Vậy ........

\(pt\text{⇔}x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\text{⇔}6x+20=0\text{⇔}x=-\dfrac{10}{3}\)

Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

đặt \(t=x^2\) (t\(\ge\)0) pt thành \(t^2-3t-28\)=0<=>(t-7)(t+4)=0

=>\(\left[{}\begin{matrix}t=7\\t=-4\end{matrix}\right.\)so với điều kiện=>t=7

=>\(x^2=7\)<=>\(\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\)

KL: S=\(\pm\sqrt{7}\)

x^4-3x^2-28=0

=>x^4 - 7x^2 + 4x^2 - 28=0

=>x^2(x^2 - 7) + 4(x^2 - 7)=0

=>(x^2 - 7)(x^2 + 4)=0

=>x^2 - 7=0  hoặc  x^2 + 4=0

=>x^2=7   hoặc   x^2= -4

=>x=căn 7     hoặc     không có x thỏa mãn

KẾT LUẬN

Mình không chắc lắm, nếu có sai thì cho mình xin lỗi nha

CHÚC BẠN HỌC TÔT

bn tham khảo https://hoc247.net/hoi-dap/toan-8/tim-x-biet-x-2-x-2-4-x-2-x-12-0-faq378404.html

bạn viết lại đề đi

đề là tìm x ạ

ĐKXĐ: \(x\ge2\)

\(\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=5\)

- TH1:

\(2\le x\le6\)

\(\Rightarrow2-\sqrt{x-2}+3-\sqrt{x-2}=5\)

\(\Leftrightarrow\sqrt{x-2}=0\Rightarrow x=2\)

TH2: \(6\le x\le11\)

\(\Rightarrow\sqrt{x-2}-2+3-\sqrt{x-2}=5\Leftrightarrow1=5\) (vô nghiệm)

TH3: \(x>11\)

\(\Rightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=5\)

\(\Leftrightarrow2\sqrt{x-2}=10\Rightarrow\sqrt{x-2}=5\)

\(\Rightarrow x=27\)

Vậy \(\left[{}\begin{matrix}x=2\\x=27\end{matrix}\right.\)

Đk: \(x\ge4\)

\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

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

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

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

TH1:\(\sqrt{x-4}>2\Leftrightarrow x>8\)

\(A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

TH2:\(\sqrt{x-4}\le2\Leftrightarrow4\le x\le8\)

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

Vậy...

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)