a ) Nếu \(x=71\) \(\Rightarrow70=x-1\)

Thay \(70=x-1\) vào A , ta được :

\(A=x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)

\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x\)

\(=x\)

\(=71\)

Vậy \(A=71\) tại \(x=71\)

b ) Ta có : \(x=35\)

\(\Rightarrow\left\{{}\begin{matrix}36=x+1\\37=x+2\\69=2x-1\\34=x-1\end{matrix}\right.\) ( * )

Thay ( * ) vào B , ta được :

\(B=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)

\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)

\(=x+15\)

\(=35+15=50\)

Vậy \(B=50\) tại \(x=35\)

Nhầm mọi người giúp mk nhoa

\(a,\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\\ =\dfrac{3\times4\times2\times15}{2\times5\times3\times4}\\ =\dfrac{3\times4\times2\times5\times3}{2\times5\times3\times4}\\ =3\)

\(b,\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\\ =\dfrac{6\times5\times7\times7\times8}{7\times8\times3\times6\times5}\\ =\dfrac{7}{3}\)

a) \(\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\)

\(=\left(\dfrac{3}{2}\times\dfrac{2}{3}\right)\times\left(\dfrac{4}{5}\times\dfrac{15}{4}\right)\)

\(=1\times3\)

\(=3\)

b) \(\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\)

\(=\left(\dfrac{6}{7}\times\dfrac{7}{6}\right)\times\left(\dfrac{5}{8}\times\dfrac{8}{5}\right)\times\dfrac{7}{3}\)

\(=1\times1\times\dfrac{7}{3}\)

\(=\dfrac{7}{3}\)

Chúc bạn học tốt

Ta có:

\(x=35\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\x+2=27\\2x-1=69\\x-1=34\end{matrix}\right.\) (1)

Thay (1) vào biểu thức ta được:

\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)

\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)

\(=x+15\)

\(=35+15\)

\(=50\)

50

a)\(A=x^5-36x^4+37x^3-69x^2+34x+15\)

=\(x^5-35x^4-x^4+35x^3+2x^2-70x^2+x^2-35x+x+15\)

=\(\left(x^4-x^3+x^2+x\right)\left(x-35\right)+x+15\)

=0+35+15=50(do x=35)

b, \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)

\(\Rightarrow3\left(24x^2+6x-20x-5\right)-\left(72x^2-16x+27x-6\right)=203\)

\(\Rightarrow72x^2-42x-15-72x^2-11x+6=203\)

\(\Rightarrow-53x=203-6+15=212\)

\(\Rightarrow x=-4\)

Chúc bạn học tốt!!!

x^5 - (x-1)x^4 -(x-1)x^3 - (x-1)x^2 - (x-1)x + 34
= x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x + 34 
= 71 + 34 = 105

\(\dfrac{11}{34}+\dfrac{5}{23}+\dfrac{23}{34}+\dfrac{18}{23}\)
\(=\left(\dfrac{11}{34}+\dfrac{23}{34}\right)+\left(\dfrac{5}{23}+\dfrac{18}{23}\right)\)
\(=1+1\)
\(=2\)
#DatNe

=2

\(\dfrac{11}{34}+\dfrac{5}{23}+\dfrac{23}{34}+\dfrac{18}{23} =\left(\dfrac{11}{34}+\dfrac{23}{34}\right)+\left(\dfrac{5}{23}+\dfrac{18}{23}\right)=\dfrac{34}{34}+\dfrac{23}{23}=1+1=2\)