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

Đặt y=x+1

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

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

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

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

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

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

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

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

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