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a) Ta có : \(x=31\Rightarrow30=x-1\)
Thay vào biểu thức ta được:
\(A=x^3-\left(x-1\right).x^2-x^2+1=x^3-x^3+x^2-x^2+1=1\)
b) Ta có: \(x=9\Rightarrow x+1=10\)
Thay vào biểu thức ta được
\(B=x^{14}-\left(x+1\right).x^{13}+\left(x+1\right).x^{12}-\left(x+1\right).x^{11}+.....+x^2.\left(x+1\right)=\left(x+1\right).x+\left(x+1\right)\)
\(\Leftrightarrow B=x^{14}-x^{14}-x^{13}+x^{13}+....+x^3+x^2=x^2+2x+1\)
\(\Leftrightarrow B=x^2-x^2-2x-1=-2.9-1=-19\)
\(x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
a) Ta có: \(A=x^5-15x^4+16x^3-29x^2+13x\)
\(=\left(x^5-14x^4\right)-\left(x^4-14x^3\right)+\left(2x^3-28x^2\right)-\left(x^2-14x\right)-x\)
\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-x\)
\(=\left(x-14\right)\left(x^4-x^3+2x^2-x\right)-x\)(thay x = 14)
\(=-x=-14\)
Vậy A = -14.
b) Ta có: \(B=x^{14}-10x^3+10x^{12}-10x^{11}+...+10x^2-10x+10\) tại x = 9.
\(\cdot x=9\Rightarrow10=x+1\)
\(\Rightarrow B=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{13}-x^{12}+...+x^3+x^2-x^2-x+10\)
\(=-x-10=-9-10=-19.\)
Vậy B = -19.
a) Ta có:
\(A=x^5-15x^4+16x^3-29x^2+13x\)
\(=\left(x^5-14x^4\right)-\left(x^4-14x^3\right)+\left(2x^3-28x^2\right)-\left(x^2-14x\right)-x\)
\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-x\)
\(=\left(x-14\right)\left(x^4-x^3+2x^2-x\right)-x\)(thay \(x=14\))
\(=-x=-14\)
Vậy \(A=-14\)
b) Ta có:
\(B=x^{14}-10x^3+10x^{12}-10x^{11}+...+10x^2-10x+10\)tại \(x=9\)
\(x=9\Rightarrow10=x+1\)
\(\Rightarrow B=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{13}-x^{12}+...+x^3+x^2-x^2-x+10\)
\(=-x-10=-9-10=-19\)
Vậy \(B=-19\)
x=9=>10=x+1
thqy 10=x+1 vào A
ta có A=x^14 - (x+1)x^13+(x+1)x^12-(x+1)x^11+...+(x+1)x^2-(x+1)x+10
=x^14-x^14-x^13+x^13+x^12-x^12-x^11+...+x^3+x^2-x^2_x+10
=x+10
mà x=9
=>A=19
\(.\)M= bn ghi lại đề nha ^.^
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=1^3-3ab.1+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2.1\)
\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)
\(M=1-3ab+3ab-6a^2b^2+6a^2b^2\)\(=1\)
k cho mình nha bn thanks nhìu <3 <3 (^3^)
2. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(1)
Đặt \(x^2+5x+4=t\)
(1) = \(t.\left(t+2\right)-24\)
\(=t^2+2t+1-25\)
\(=\left(t+1\right)^2-25\)
\(=\left(t+1-5\right)\left(t+1+5\right)\)
\(=\left(t-4\right)\left(t+6\right)\)(2)
Thay \(t=x^2+5x+4\)vào (2) ta có:
(2) = \(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
k mình nha bn <3 thanks
a) \(C=x^3+3x^2+3x+10=\left(x+1\right)^3+9\)
Tại x = 99...9 (2004 chữ số 9) thì: x+1 = 100...0 (2004 chữ số 0) = 102004
Khi đó, C = (102004)3 + 9 = 106012 + 9.
b) \(B=\left(5x-11\right)^2-\left(10x-22\right)\left(5x-9\right)+\left(5x-9\right)^2=\)
\(=\left(5x-11\right)^2-2\cdot\left(5x-11\right)\left(5x-9\right)+\left(5x-9\right)^2=\left(5x-11-\left(5x-9\right)\right)^2=\left(-2\right)^2=4\)
Hay B = 4 với mọi x .
Vậy tại x = 20052006 thì B = 4.
\(A=x^{14}-10x^{13}+10x^2-10x^{11}\)\(+...+10x^{12}-10x+10\)
Thay x = 9 vào biểu thức A
\(\Rightarrow A=9^{14}-\left(9+1\right).9^{13}+\left(9+1\right).9^{12}\)\(-...+9+1\)
\(\Rightarrow A=9^{14}-9^{14}-9^{13}+9^{12}+...-9+9+1\)
\(\Rightarrow A=1\)
P/s tham khảo thêm trên google
\(15\left(2a^2-1\right)+5\left(3-\frac{1}{5a}-6a^2\right)\)
\(=30a^2-15+15-\frac{1}{a}-30a^2\)
\(=-\frac{1}{a}\)
tại \(a=2017\)=> M= \(\frac{-1}{a}=\frac{-1}{2017}\)
\(\left(x-y\right)\left(x^2+xy+y^2\right)+y^3\)
\(=x^3-y^3+y^3\)
\(=x^3\)
ại \(x=2\)=> N= \(x^3=2^3=8\)
(Câu trả lời bằng hình ảnh)
thế a vào