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)

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

Đặt: \(t=x-\frac{5+2}{2}=x-\frac{7}{2}\)

pt trở thành: \(\left(t+\frac{7}{2}-5\right)^4+\left(x+\frac{7}{2}-2\right)^4=17\)

<=> \(\left(t-\frac{3}{2}\right)^4+\left(t+\frac{3}{2}\right)^4=17\)  ( Nếu em nhớ hằng đẳng thức (a+b)^4 thì có thể làm tắt rồi rút gọn )

<=> \(\left[\left(t-\frac{3}{2}\right)^2+\left(t+\frac{3}{2}\right)^2\right]^2-2\left(t-\frac{3}{2}\right)^2\left(t+\frac{3}{2}\right)^2=17\)

<=> \(\left(2t^2+\frac{9}{2}\right)^2-2\left(t^2-\frac{9}{4}\right)^2=17\)

<=> \(2t^4+27t^2-\frac{55}{8}=0\)

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

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

<=> \(\orbr{\begin{cases}t^2=\frac{1}{4}\\t^2=-\frac{55}{4}\left(loai\right)\end{cases}}\Leftrightarrow t=\pm\frac{1}{2}\)

Với  t = 1/2 em thay vào tính x

       t =-1/2 ....

4(x – 4) = -7x +17

⇔ 4x - 16 = -7x + 17

⇔ 4x + 7x = 17 + 16

⇔ 11x = 33

⇔ x = 3

Phương trình có tập nghiệm S = { 3}

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)

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

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

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

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

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

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

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

=.= hok tốt!!

\(\frac{x-17}{1990}+\frac{x-21}{1986}+\frac{x+1}{1004}=4\)

\(\Leftrightarrow\left(\frac{x-17}{1990}-1\right)+\left(\frac{x-21}{1986}-1\right)+\left(\frac{x+1}{1004}-2\right)=0\)

\(\Leftrightarrow\frac{x-2007}{1990}+\frac{x-2007}{1986}+\frac{x-2007}{1004}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{1990}+\frac{1}{1986}+\frac{1}{1004}\right)=0\)

\(\Leftrightarrow x-2007=0\) (Vì \(\frac{1}{1990}+\frac{1}{1986}+\frac{1}{1004}>0\))

\(\Leftrightarrow x=2007\)

V...

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

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

\(\frac{x-17}{33}+\frac{169-x}{23}+\frac{x}{25}=4\)

\(\Rightarrow575.\left(x-17\right)+825.\left(169-x\right)+759x=75900\)

\(\Rightarrow575x-9775+139425-825x+759x-75900=0\)

\(\Rightarrow509x=-53750\)

\(\Rightarrow x=\frac{-53750}{509}\)

sử dụng tỉ lệ con nhà bà thức ta có (:|

\(\Leftrightarrow\frac{509x+129650}{18975}=\frac{4}{1}\Rightarrow\left(509x+129650\right)1=18975.4\)

\(\Rightarrow\frac{\left(509x+129650\right)1}{509x}=\frac{18975.4}{509x}\)

\(\Rightarrow\frac{509x+129650}{509x}=\frac{18975.4}{509x}\)

\(\Rightarrow x=-105,599214145383\)