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x=7=>x+1=8
A=x15-(x+1)14+(x+1)x13-(x+1)x12+...-(x+1)x2+(x+1)x-5=x15-x15-x14+x14+x13-x13-x12+...-x3-x2+x2+x-5=7-5=2
Bài 1:
\(f\left(x\right)=-x^{15}+8x^{14}-8x^{13}+...-8x-5\)
Ta xét \(x=7\Leftrightarrow x+1=8\)
Khi đó :
\(f\left(7\right)=-x^{15}+x^{14}\left(x+1\right)-x^{13}\left(x+1\right)+...-x\left(x+1\right)-5\)
\(f\left(7\right)=-x^{15}+x^{15}+x^{14}-x^{14}-x^{13}+...-x^2-x-5\)
\(f\left(7\right)=-x-5\)
\(f\left(7\right)=-7-5\)
\(f\left(7\right)=-12\)
Vậy...
Ta có:
x=7=>x+1=8
A=x15-(x+1)14+(x+1)x13-(x+1)x12+...-(x+1)x2+(x+1)x-5=x15-x15-x14+x14+x13-x13-x12+...-x3-x2+x2+x-5=7-5=2
Vậy A=2
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
Ta có : \(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\frac{\left(\sqrt{5}+2\sqrt[3]{\sqrt{5}-2^{ }}\right)^3}{\sqrt{5}+3-\sqrt{5}}\) 2)3 trong căn bậc nhé mk ko vt đc ( ko bt giải thick thông cảm )
\(=\frac{\sqrt{5}^2-2^2}{3}\)
\(=\frac{1}{3}\)
Vậy \(A=\left(3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2\right)^{2011}=3^{2011}\)
Trả lời
A=(3x3+8x2+2)2011 với x=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}\sqrt{9-6\sqrt{5}+5}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}\)
=\(\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{3}\)
=1/3
Học tốt !
Bài 1:
a) Ta có: \(x=7\Rightarrow8=x+1\)
Thay vào ta được:
\(A=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(A=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)
\(A=x-5\)
\(A=7-5=2\)
Vậy khi x = 7 thì A = 2
x = 7 => x + 1 = 8
=> Biểu thức có giá trị là:
x15 - (x+1).x14 + (x+1).x13 - ... + (x+1).x - 5
= x15 - x15 - x14 + x14 + x13 - ... + x2 + x - 5
= x - 5
= 7 - 5
= 2