Câu 1:

Đặt \(x+1=a\). Khi đó \(x+3=a+2; x-1=a-2\).

PT đã cho tương đương với:

\((a+2)^4+(a-2)^4=626\)

\(\Leftrightarrow 2a^4+48a^2+32=626\)

\(\Leftrightarrow a^4+24a^2-297=0\)

\(\Leftrightarrow (a^2+12)^2=441\)

\(\Rightarrow a^2+12=\sqrt{441}=21\) (do \(a^2+12>0)\)

\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)

Nếu $a=3$ thì \(x=a-1=2\)

Nếu $a=-3$ thì $x=a-1=-4$

Câu 2:

Đặt \(2x-1=a; x-1=b\). PT đã cho tương đương với:

\(a^3+b^3+(-a-b)^3=0\)

\(\Leftrightarrow a^3+b^3-(a+b)^3=0\)

\(\Leftrightarrow a^3+b^3-[a^3+b^3+3ab(a+b)]=0\)

\(\Leftrightarrow ab(a+b)=0\Rightarrow \left[\begin{matrix} a=0\\ b=0\\ a+b=0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-1=0\\ 3x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\\ x=\frac{2}{3}\end{matrix}\right.\)

đặt x+2=a ,,lm típ đi,,mik ủng hộ

Sửa đề: \(\left(4,5-x\right)^4+\left(5,5-x\right)^4=1\)

=>\(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)

=>\(\left\lbrack\left(x-5\right)+0,5\right\rbrack^4+\left\lbrack\left(x-5\right)-0,5\right\rbrack^4=1\) (1)

Đặt a=x-5; b=0,5

\(\left(a+b\right)^4+\left(a-b\right)^4\)

\(=\left(a^2+2ab+b^2\right)^2+\left(a^2-2ab+b^2\right)^2\)

\(=\left\lbrack\left(a^2+b^2\right)^2+2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack+\left\lbrack\left(a^2+b^2\right)^2-2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack\)

\(=2\left(a^4+2a^2b^2+b^4\right)+8a^2b^2=2a^4+12a^2b^2+2b^4\)

\(=2\left(x-5\right)^4+12\cdot\left(x-5\right)^2\cdot\left(0,5\right)^2+2\cdot\left(0,5\right)^4\)

\(=2\left(x-5\right)^4+3\left(x-5\right)^2+0,125\)

(1) sẽ trở thành: \(2\left(x-5\right)^4+3\left(x-5\right)^2+0,125=1\)

=>\(2\left(x-5\right)^4+3\left(x-5\right)^2-0,875=0\)

=>\(16\left(x-5\right)^4+24\left(x-5\right)^2-7=0\)

=>\(16\left(x-5\right)^4+28\left(x-5\right)^2-4\left(x-5\right)^2-7=0\)

=>\(\left\lbrack4\left(x-5\right)^2+7\right\rbrack\left\lbrack4\left(x-5\right)^2-1\right\rbrack=0\)

=>\(4\left(x-5\right)^2-1=0\)

=>\(\left(2x-10\right)^2=1\)

=>\(\left[\begin{array}{l}2x-10=1\\ 2x-10=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=11\\ 2x=9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}\\ x=\frac92\end{array}\right.\)

Áp dụng bảng tam giác Pascal ta có : 

\(\left(x-2\right)^4=x^4-8x^3+24x^2-32x+16\)

\(\left(x+2\right)^4=x^4+8x^3+24x^2+32x+16\)

\(\Rightarrow\left(x-2\right)^4+\left(x+2\right)^4=2x^4+48x^2+32=626\)

\(\Leftrightarrow2x^4+48x^2-594=0\)

\(\Leftrightarrow2x^4-6x^3+6x^3-18x^2+66x^2-594=0\)

\(\Leftrightarrow2x^3\left(x-3\right)+6x^2\left(x-3\right)+66\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(2x^3+6x^2+66x+198\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[2x^2\left(x+3\right)+66\left(x+3\right)\right]\left(x-3\right)=0\)

\(\Leftrightarrow2\left(x+3\right)\left(x^2+33\right)\left(x-3\right)=0\)

\(\Rightarrow x=\pm3\)

Vậy nghiệm \(S=\left\{\pm3\right\}\)

e:

Tham khảo: 

a: \(\Leftrightarrow x^2-2x+1+4x^2+4x+4-5x^2+5=0\)

\(\Leftrightarrow2x+10=0\)

hay x=-5