Thay x+1=80 ta đc:

\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)

\(=x^7-x^7-x^6+x^6+x^5+...+x^2+x+15\)

\(79+15=94\)

\(Ta \)  \(có \) \(:\)

\(x = 79 \)\(\Rightarrow\)\(x + 1 = 80\)

\(Thay \)  \(x + 1 = 80 \) \(vào \)  \(P(x)\) \(ta\) \(được :\)

\(P ( x ) = x ^7 - ( x + 1 )x ^6 + ( x + 1 )x^5\)\(- ( x + 1 )x ^4\)\(+ ...+ ( x + 1 )x + 15\)

\(P ( x ) = x ^7 - x ^7- x^6 + x^6 + x^5 - x^ 5\)\(- x ^4 + x ^4 + ... - x^ 2 + x ^2 + x + 15\)

\(P ( x ) = x + 15\)

\(Thay x = 79 vào P ( x ) ta được :\)

\(P ( x ) = 79 + 15 = 94\)

(156 + 78) x 6 ... 156 x 6 + 79 x 6

=> (156 + 78) x 6 .... 6 x (156 +79)

Vì 156 + 78 < 156 + 79

=> (156 + 78) x 6 > 156 x 6 +79 x 6

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\(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4\)\(+...+\left(x+1\right)x+15\)

\(P\left(x\right)=x^7-x^7-x^6+x^6+...+x^2+x+15\)

\(P\left(x\right)=x+15=94\)

Vậy giá trị của P(x) tại x = 79 là 94 

ng qua ko bt lam a bn

Nguyễn Khắc Nam m khủng hả con

\(\frac{4x}{7}=\frac{5y}{11}=\frac{4x+5y}{7+11}=\frac{4\left(x+y\right)+y}{18}=\frac{4.79+y}{18}.\)

\(\Rightarrow\frac{5y}{11}=\frac{4.79+y}{18}\Rightarrow90y=4.11.79+11y\)

\(\Rightarrow79y=4.11.79\Rightarrow y=44\)

\(\frac{4x}{7}=\frac{5y}{11}=\frac{5.44}{11}=20\Rightarrow x=35\)

(10 - 1 \(\dfrac{1}{6}\) x \(\dfrac{6}{7}\)) : ( 21: \(\dfrac{11}{2}\) + 5\(\dfrac{2}{11}\) ) 

( 10 \(\dfrac{7}{6}\) x \(\dfrac{6}{7}\) ) : ( 21 : \(\dfrac{11}{2}\) + \(\dfrac{57}{11}\) )

(10 - 1) : ( \(\dfrac{42}{11}\) + \(\dfrac{57}{11}\) )

= 9 : 9

= 1

=> đáp án : 1

=(10 - 1 1/6 x 6/7) : ( 21: 11/2 + 5 2/11) 

=( 10 7/6 x 6/7 ) : ( 21 : 11/2 + 57/11)

=(10 - 1) : ( 42/11 + 57/11 )

= 9 : 9

=1

=(10 - 1 1/6 x 6/7) : ( 21: 11/2 + 5 2/11) 

=( 10 7/6 x 6/7 ) : ( 21 : 11/2 + 57/11)

=(10 - 1) : ( 42/11 + 57/11 )

= 9 : 9

=1