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Mk mak lm mấy cái con toán này
Thì mk sáng mai cx ko lm xng đôu
Bấm mẹ máy tính cho nó nhanh
K mk
\(a)\)\(\frac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\)\(\frac{2^{12}.5^{12}+2^9.5^{11}-2^8.5^{13}}{2^8.5^{11}}\)
\(=\)\(\frac{2^8.5^{11}\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=\)\(2^4.5+2-5^2\)
\(=\)\(80+2-25\)
\(=\)\(57\)
Chúc bạn học tốt ~
\(\left(10^{12}+5^{11}.2^9-5^{13}.2^8\right):4.5^5.10^6\)
\(=\left(5^{12}.2^{12}+5^{11}.2^9-5^{13}.2^8\right):2^2.5^5.2^6.5^6\)
\(=\left[5^{11}.2^8\left(5.2^4+2-5^2\right)\right]:2^8.5^{11}\)
\(=\frac{5^{11}.2^8\left(5.16+2-25\right)}{5^{11}.2^8}\)
\(=80+2-25\)
\(=57\)
\(\frac{3^{30}-5.3^{16}.3^{12}+4.3^{16}.3^8}{41.3^{24}}=\frac{3^{24}\left(3^6-5.3^4+4\right)}{41.3^{24}}=\frac{3^4\left(3^{2-5}\right)+4}{41}=\frac{4\left(3^4+1\right)}{41}=\frac{4.82}{41}=8\)
a, x = 79 => x + 1 = 80
Ta có:\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)
\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+x^2+x+15\)
\(=x+15=79+15=94\)
Còn lại tương tự
\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
\(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
Lời giải:
a) Với \(x=79\)
\(P(x)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=(x^7-79x^6)-(x^6-79x^5)+(x^5-79x^4)-....-(x^2-79x)+x+15\)
\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-...-x(x-79)+x+15\)
\(=(x^6-x^5+x^4-...-x)(x-79)+x+15\)
\(=(x^6-x^5+x^4-...-x)(79-79)+79+15=79+15=94\)
b) Hoàn toàn tương tự phần a.
\(Q(x)=(x^{14}-9x^{13})-(x^{13}-9x^{12})+(x^{12}-9x^{11})-...+(x^2-9x)-x+10\)
\(=x^{13}(x-9)-x^{12}(x-9)+x^{11}(x-9)-...+x(x-9)-x+10\)
\(=(x-9)(x^{13}-x^{12}+x^{11}-...+x)-x+10\)
\(=(9-9)(x^{13}-x^{12}+...+x)-9+10=0-9+10=1\)
c)
\(R(x)=(x^4-16x^3)-(x^3-16x^2)+(x^2-16x)-x+20\)
\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)
\(=(x-16)(x^3-x^2+x)-x+20\)
Với $x=16$ thì $Q(x)=(16-16)(x^3-x^2+x)-16+20=0-16+20=4$
d)
\(S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+x(x-12)-x+10\)
\(=x^9(x-12)-x^8(x-12)+x^7(x-12)-...+x(x-12)-x+10\)
\(=(x-12)(x^9-x^8+x^7-..+x)-x+10\)
\(=(12-12)(x^9-x^8+x^7-...+x)-12+10=-12+10=-2\)
CẢM ƠN NHÌU
1. 8 - 12x + 6x2 - x3
= 23 - 3.22.x + 3.x2.2 - x3
=(2-x)3
2. 125x3 - 75x2 +15x - 1
=(5x)3 - 3.(5x)2.1 + 3.5x.12 - 13
=(5x - 1)3
3, 4 (sai đề)
5. x3 + 2x2 - 6x - 27
=(x3 - 27) + (2x2 - 6x)
=(x3 - 33) + (2x2 - 6x)
=(x -3)(x2 + 3x + 9) + 2x(x-3)
=(x-3)(x2 + 3x +9 +2x)
=(x-3)(x2 + 5x +9)
6. 12x3 + 4x2- 27x -9
=(12x3 + 4x2) - (27x + 9)
=4x2(3x + 1) - 9(3x +1)
=(3x -1)(4x2 -9)
=(3x-1)(2x-3)(2x+3)
C1 : lần trước mình giải
C2 : mình không chắc thử xem
thay x= 9 vao F ta có
F = 9^14 - 10 .9^13 + 10.9^12 - 10 .9^11 + ... +10.9^2 -10.9 + 10
= 9^14 - ( 9 + 1 ) . 9^13 + (9+1). 9^12+..+(9+1) .9^2 - (9+1)9 +10
= 9^14 - 9^14 - 9^13 + 9^13 + 9^12 -.....+9^3 + 9^2 - 9^2 - 9 + 10 = 1
Tương tự vói G , H
C gửi tích giúp c nha
