Câu hỏi của Nguyễn Thanh Xuân

Question

tính giá trị của biểu thức a) A= x5  - 15x4 + 16x3 - 29x2 +13x        tại x = 4b) B= x14  - 10x13 + 10x12 - 10x11 + ..... + 10x2 - 10x +10       tại x=9

Lê Tuấn Kiệt · Accepted Answer

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.Câu trả lời: 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.

Doraemon · Answer

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$$=-x-10=-9-10=-19$Vậy $B=-19$Câu trả lời: a) Ta có:$A=x^5-15x^4+16x^3-29x^2+13x$$=\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 $B=-19$

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