Ta có: \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)

<=> \(x=9-\left(\dfrac{\sqrt{\dfrac{9}{4}+\sqrt{5}}-\sqrt{\dfrac{9}{4}-\sqrt{5}}}{\left(\sqrt{\dfrac{9}{4}-\sqrt{5}}\right)\left(\sqrt{\dfrac{9}{4}}+\sqrt{5}\right)}\right)\)

<=> \(x=9-\left(\dfrac{\sqrt{\dfrac{9}{4}+\sqrt{5}}-\sqrt{\dfrac{9}{4}-\sqrt{5}}}{\sqrt{\dfrac{81}{16}-5}}\right)\)

<=> \(x=9-\left(\dfrac{\sqrt{\dfrac{9}{4}+\sqrt{5}}-\sqrt{\dfrac{9}{4}-\sqrt{5}}}{\dfrac{1}{4}}\right)\)

Đặt \(D=\sqrt{\dfrac{9}{4}+\sqrt{5}}-\sqrt{\dfrac{9}{4}-\sqrt{5}}\)

<=> \(D^2=\left(\sqrt{\dfrac{9}{4}+\sqrt{5}}-\sqrt{\dfrac{9}{4}-\sqrt{5}}\right)^2\)

\(=\dfrac{9}{4}+\sqrt{5}+\dfrac{9}{4}-\sqrt{5}-2\sqrt{\left(\sqrt{\dfrac{9}{4}+\sqrt{5}}\right)\left(\sqrt{\dfrac{9}{4}-\sqrt{5}}\right)}\)

<=> \(D^2=\dfrac{9}{2}-2.\sqrt{\dfrac{1}{16}}=\dfrac{9}{2}-2.\dfrac{1}{4}=4\)

<=> \(D=\sqrt{4}=2\)

=> \(x=9-\dfrac{2}{\dfrac{1}{4}}=1\)

Mà \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\)

=> \(f\left(1\right)=\left(1-3+1\right)^{2016}=1\)

Hay \(f\left(x\right)=1\) khi \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)

P/s: Đã lm chậm nhất có thể!

Sửa đề: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)

\(\Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\cdot\sqrt{x-5}-\dfrac{1}{5}\cdot5\sqrt{x-5}=6\)

\(\Leftrightarrow\sqrt{x-5}=6\)

=>x-5=36

hay x=41

\(\dfrac{5}{7}x-x=1\dfrac{1}{7}\\ < =>\dfrac{5}{7}x-x=\dfrac{8}{7}\\ < =>\left(\dfrac{5}{7}-\dfrac{7}{7}\right)x=\dfrac{8}{7}\\ < =>-\dfrac{2}{7}x=\dfrac{8}{7}\\ =>x=\dfrac{\dfrac{8}{7}}{-\dfrac{2}{7}}=-4\)

=-4

\(PT\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\dfrac{13}{9}\)

Vậy: \(S=\left\{\dfrac{13}{9}\right\}\)

\(5-\left(6-x\right)=4\left(3-2x\right)\)

\(5-6+x=12-8x\)

\(-1+x=12-8x\)

\(x-1=12-8x\)

\(12+1=8x+1\)

\(8x=13-1\)

\(x=12:8\)

\(x=\dfrac{12}{8}=\dfrac{3}{2}\)

Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:

ĐKXĐ: ...

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)

Đặt \(4x+\frac{7}{x}-10=t\)

\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

<=> (x-1-3x-3)(x-1+3x+3) = 0

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)

\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)

`3/4 + 5/6 = 9/12 + 10/12 = 19/12`

`1/2 + 7/12 = 6/12 + 7/12 = 13/12`

`2/3 xx 3/4 = 2/4 = 1/2`

`7/4 : 2 = 7/4 xx 1/2 = 7/8`

\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)

\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)

\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)

\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)

`9/14 : 3/7 + 5/6 = 3/98 + 5/6 = 127/147`

__________________

`4/7 xx (4 - 7/3) = 4/7 xx 5/3 = 20/21`

\(\dfrac{9}{14}:\dfrac{3}{7}+\dfrac{5}{6}=\dfrac{9}{14}\times\dfrac{7}{3}+\dfrac{5}{6}=\dfrac{3}{2}+\dfrac{5}{6}=\dfrac{7}{3}\)

\(\dfrac{4}{7}\times\left(4-\dfrac{7}{3}\right)=\dfrac{4}{7}\times\dfrac{5}{3}=\dfrac{20}{21}\)

\(=\dfrac{3\times3}{2\times7}\times\dfrac{7}{3}+\dfrac{5}{6}=\dfrac{3}{2}+\dfrac{5}{6}=\dfrac{9}{6}+\dfrac{5}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

b)\(=\dfrac{4}{7}\times\dfrac{5}{3}=\dfrac{20}{21}\)