a, (4 - x )⁵ +(x - 2)⁵ =32

(=) 1024  -  x⁵ + x⁵  - 32 = 32

(=) -x⁵ + x⁵ = 32 + 32 - 1024

(=) 0x =  -960

=) phương trình vô nghiệm

sauriengroblox

3) (x−1)5+(x+3)5=242(x+1)(x−1)5+(x+3)5=242(x+1)

- Đặt x+1=y⇒x−1=y−2;x+3=y+2x+1=y⇒x−1=y−2;x+3=y+2

- Ta có:

(y−2)5+(y+2)5=242y(y−2)5+(y+2)5=242y

⇔y5−5y4+40y3−80y2+80y−32+y5+5y4+40y3+80y2+80y+32=242y⇔y5−5y4+40y3−80y2+80y−32+y5+5y4+40y3+80y2+80y+32=242y

⇔2y5+80y3+160y−242y=0⇔2y5+80y3+160y−242y=0

⇔2y5+80y3−82y=0⇔2y5+80y3−82y=0

⇔2y(y4+40y2−41)=0⇔2y(y4+40y2−41)=0

⇔2y(y4+41y2−y2−41)=0⇔2y(y4+41y2−y2−41)=0

⇔2y[y2(y2+41)−(y2+41)]=0⇔2y[y2(y2+41)−(y2+41)]=0

⇔2y(y2−1)(y2+41)=0⇔2y(y2−1)(y2+41)=0

⇔2y(y−1)(y+1)(y2+41)=0⇔2y(y−1)(y+1)(y2+41)=0

⇔y=0;y−1=0⇒y=1;y+1=0⇒y=−1⇔y=0;y−1=0⇒y=1;y+1=0⇒y=−1

⇔y=0⇒x+1=0⇒x=−1⇔y=0⇒x+1=0⇒x=−1

⇔y=1⇒x+1=1⇒x=0⇔y=1⇒x+1=1⇒x=0

⇔y=−1⇒x+1=−1⇒x=−2

\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

Chắc Sai kết quả chứ công thức đúng nha!!!...

Fighting!!!...

Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)

Đặt y=x+1

\(\left(x-1\right)^5=\left(y-2\right)^5\)

\(=y^5-5y^4+40y^3-80y^2+80y-32\)

\(\left(y+2\right)^5=y^5+5y^4+40y^3+80y^2+80y+32\)

\(A=\left(y+2\right)^5+\left(y-2\right)^5-242y\)

\(=2y^4+80y^3+160y-242y\)

\(=2y^4+80y^3-82y\)

\(=2y\left(y^2-1\right)\left(y^2+41\right)\)

\(=2y\left(y+1\right)\left(y-1\right)\left(y^2+41\right)\)

\(=2\left(x+1\right)\left(x+2\right)\cdot x\cdot\left[\left(x+1\right)^2+41\right]\)