\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

\(a.P(x)=x^7-80x^6+80x^5-80x^4+....+80x+15\)

\(=x^7-79x^6-x^6+79x^5+x^5-79x^4-....-x^2+79x+x+15\)

\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-....-x(x-79)+x+15\)

\(=(x-79)(x^6-x^5+x^4-....-x)+x+15\)

Thay x = 79 vào biểu thức trên , ta có

\(P(79)=(79-79)(79^6-79^5+79^4-...-79)+79+15\)

\(=0+79+15\)

\(=94\)

Vậy \(P(x)=94\)khi x = 79

\(b.Q(x)=x^{14}-10x^{13}+10x^{12}-.....+10x^2-10x+10\)

\(=x^{14}-9x^{13}-x^{13}+9x^{12}+.....-x^3+9x^2+x^2-9x-x+10\)

\(=x^{13}(x-9)-x^{12}(x-9)+.....-x^2(x-9)+x(x-9)-x+10\)

\(=(x-9)(x^{13}-x^{12}+.....-x^2+x)-x+10\)

Thay x = 9 vào biểu thức trên , ta có

\(Q(9)=(9-9)(9^{13}-9^{12}+.....-9^2+9)-9+10\)

\(=0-9+10\)

\(=1\)

Vậy \(Q(x)=1\)khi x = 9

\(c.R(x)=x^4-17x^3+17x^2-17x+20\)

\(=x^4-16x^3-x^3+16x^2+x^2-16x-x+20\)

\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)

\(=(x-16)(x^3-x^2+x)-x+20\)

Thay x = 16 vào biểu thức trên , ta có

\(R(16)=(16-16)(16^3-16^2+16)-16+20\)

\(=0-16+20\)

\(=4\)

Vậy \(R(x)=4\)khi x = 16

\(d.S(x)=x^{10}-13x^9+13x^8-13x^7+.....+13x^2-13x+10\)

\(=x^{10}-12x^9-x^9+12x^8+.....+x^2-12x-x+10\)

\(=x^9(x-12)-x^8(x-12)+....+x(x-12)-x+10\)

\(=(x-12)(x^9-x^8+....+x)-x+10\)

Thay x = 12 vào biểu thức trên , ta có

\(S(12)=(12-12)(12^9-12^8+....+12)-12+10\)

\(=0-12+10\)

\(=-2\)

Vậy \(S(x)=-2\)khi x = 12

Hình như đây là toán lớp 7 có trong phần trắc nghiệm của thi HSG huyện

Chúc bạn học tốt , nhớ kết bạn với mình

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

mn ơi giúp mình với

Híc híc

a) 7x - 2x = 6¹⁷ : 6¹⁵ + 44

=> 5x = 36 + 44

=> 5x = 80

=> x = 80 : 5 = 16

b) 9^{x - 1} = 18 + 1/9 - 1/9 - 9

=> 9^{x - 1} = 9

=> x - 1 = 1

=> x = 1 + 1 = 2

c) [(6x - 39) : 7] . 4 = 12

=> (6x - 39) : 7 = 12 : 4

=> (6x - 39) : 7 = 3

=> 6x - 39 = 3.7

=> 6x - 39 = 21

=> 6x = 21 + 39

=> 6x = 60

=> x = 60 : 6

=> x = 10

d) 2 - (x - 1) - 3x = 20

=> 2 - x + 1 - 3x = 20

=> 3 - 4x = 20

=> 4x = 3 - 20

=> 4x = -17

=> x = -17 : 4 = -17/4

e) 2|x - 3| + 7 = 5⁶ : 5²

=> 2|x - 3| + 7 = 625

=> 2|x - 3| = 625 - 7

=> 2|x - 3| = 618

=> |x - 3| = 618 : 2

=> |x - 3| = 309

=> \(\orbr{\begin{cases}x-3=309\\x-3=-309\end{cases}}\)

=> \(\orbr{\begin{cases}x=312\\x=-306\end{cases}}\)

Bài này bạn chỉ cần bỏ chút thời gian là làm đc, làm như những bài tìm x bình thường thôi

a/

\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

c, \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

A)  7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11

=7/38.(9/11+4/11-2/11)

=7/38

B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20

=5/31.(21/25-7/10-9/20)

=5/31.(-31/100)

=-1/20