a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

Ta có : 17 - 14(x + 1) = 13 - 4(x + 1) - 5(x - 3)

<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15

<=> -14x + 3 = -9x + 24

<=> -14x + 9x = 24 - 3

<=> -5x = 21

=> x = -4,2

Ta có :  5x + 3,5 + (3x - 4) = 7x - 3(x - 0,5)

<=>  5x + 3,5 + 3x - 4 = 7x - 3x + 1,5 

<=> 8x - 0,5 = 4x + 1,5

=> 8x - 4x = 1,5 + 0,5

=> 4x = 2

=> x = \(\frac{1}{2}\)

17-14(x+1)=13-4(x+1)-5(x-3)

a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)

=>9x^2-23x=x^2-5x+6

=>8x^2-18x-6=0

=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)

b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>216x+18=275x-100

=>-59x=-118

=>x=2

\(\frac{x-17}{1997}+\frac{x-21}{1993}+\frac{x+2}{1008}=4\)

\(\Leftrightarrow\frac{x-17}{1997}+\frac{x-21}{1993}+\frac{x+2}{1008}-4=0\)

\(\Leftrightarrow\left(\frac{x-17}{1997}-1\right)+\left(\frac{x-21}{1993}-1\right)+\left(\frac{x+2}{1008}-2\right)=0\)

\(\Leftrightarrow\left(\frac{x-17}{1997}-\frac{1997}{1997}\right)+\left(\frac{x-21}{1993}-\frac{1993}{1993}\right)+\left(\frac{x+2}{1008}-\frac{2016}{1008}\right)=0\)

\(\Leftrightarrow\frac{x-2014}{1997}+\frac{x-2014}{1993}+\frac{x-2014}{1008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{1997}+\frac{1}{1993}+\frac{1}{1008}\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

=.= hok tốt!!

+) <=> \(x^3-3x^2+3x-1+3x^2+6x+8-x^3=17\)

<=>9x=10

<=> x=\(\frac{10}{9}\)

+) \(x\left(x^2-25\right)-x^3-8=3\)<=> \(x^3-x^3-25x=3+8\)

<=> x=\(-\frac{11}{25}\)

\(x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\)

Ta có: \(x^2+2x+17=(x^2+2x+1)+16=\left(x+1\right)^2+16\ge16\)

\(\Rightarrow\sqrt{x^2+2x+17}\ge\sqrt{16}=4\)

\(\Rightarrow x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\ge x^4+4x^3+6x^2+4x+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1\le0\)

\(\Leftrightarrow\left(x+1\right)^4\le0\)

Mà \(\left(x+1\right)^4\ge0\Rightarrow(x+1)^4=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Thử lại ta thấy x=-1 thỏa mãn bài toán

Vậy, pt có nghiệm duy nhất là x=-1

Cho bạn kết quả phân tích thôi, tự phân tích nha:D

a) \(\Leftrightarrow2\left(x+4\right)\left(x+10\right)\left(x^2+14x+64\right)=0\)

b)\(\Leftrightarrow2\left(x-3\right)\left(x-4\right)\left(x^2-7x+26\right)=0\)

Dạng này thì em : \(\frac{6+8}{2}=7\). 

Đặt x  + 7 =t

=> Phương trình ban đầu trở thành: \(\left(t+1\right)^4+\left(t-1\right)^4=272\)

<=> \(\left(t^4+4t^3+6t^2+4t+1\right)+\left(t^4-4t^3+6t^2-4t+1\right)=272\)

<=> \(2t^4+12t^2+2=272\)

<=> \(t^4+6t^2-135=0\)

<=> \(t^4+6t^2+9=144\)

<=> \(\left(t^2+3\right)^2=12^2\)

<=> \(\orbr{\begin{cases}t^2+3=12\\t^2+3=-12\end{cases}}\Leftrightarrow\orbr{\begin{cases}t^2=9\left(tm\right)\\t^2=-15\left(l\right)\end{cases}}\Leftrightarrow t=\pm3\)

Với t = 3  có: x + 7 = 3 <=> x =-4

Với t = -3 có: x +7 =-3 <=> x = -10

b) pt  \(\left(5-x\right)^4+\left(2-x\right)^4=17\)<=> \(\left(x-5\right)^4+\left(x-2\right)^4=17\)

Tương tự: \(\frac{5+2}{2}=\frac{7}{2}\)

Đặt: \(x-\frac{7}{2}=t\)

pt trở thành: \(\left(t-\frac{3}{2}\right)^4+\left(t+\frac{3}{2}\right)^4=17\)

<=> .... 

Làm thử tiếp nha.

Chú ý công thức : \(\left(a\pm b\right)^4=a^4\pm4a^3b+6a^2b^2\pm4ab^3+b^4\)

OK!

\(\left(t-\frac{3}{2}\right)^4+\left(t+\frac{3}{2}\right)^4=17\)

<=> \(\left(t^4+4.t^3.\frac{3}{2}+6t^2.\frac{9}{4}+4t.\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4\right)\)

\(+\left(t^4-4.t^3.\frac{3}{2}+6t^2.\frac{9}{4}-4.t\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4\right)=17\)

<=> \(2t^4+27t^2-\frac{55}{8}=0\)

<=> \(t^4+\frac{27}{2}t^2-\frac{55}{16}=0\)

<=> \(\left(t^4+2.t^2.\frac{27}{4}+\frac{729}{16}\right)-\frac{729}{16}-\frac{55}{16}=0\)

<=> \(\left(t^2+\frac{27}{4}\right)^2=49\)

<=> \(t^2+\frac{27}{4}=\pm7\)

<=> \(\orbr{\begin{cases}t^2=\frac{1}{4}\\t^2=-\frac{55}{4}\left(l\right)\end{cases}}\Leftrightarrow t=\pm\frac{1}{2}\). Thay vào tìm x nhé.