tính:A=x2012+52011VỚI x+3=0
B=x2011-5x2011 +2 với x-5+0
Những câu hỏi liên quan
Cho x1+x2=x3+x4=x5+x6=…=x2011+x2012=2 và x1+x2+…+2013=0. Tìm x2013
. Cho các số dương x,y thỏa mãn :
x2010 + y2010= x2011 + y2011 = x2012 + y2012.
Tính x2016 + y2016.
\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)
\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)
\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)
\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)
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\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)
\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)
\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)
\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)
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Thực hiện phép tính:
a) \({x^5}:{x^3}\); b) \((4{x^3}):{x^2}\); c) \((a{x^m}):(b{x^n})\)(a ≠ 0; b ≠ 0; m, n \(\in\) N, m ≥ n).
a) \({x^5}:{x^3} = {x^{5 - 3}} = {x^2}\);
b) \((4{x^3}):{x^2} = (4:1).({x^3}:{x^2}) = 4x\);
c) \((a{x^m}):(b{x^n}) = (a:b).({x^m}:{x^n}) = (a:b).{x^{m - n}}\)(a ≠ 0; b ≠ 0; m, n \(\in\) N, m ≥ n).
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Tính:
a.\(\sqrt{5}-2\sqrt{20}-3\sqrt{80}\)
b.\(\dfrac{6}{\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}\)
c.\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}}{x-1}\) với x > 0;x ≠ 1
\(a,=\sqrt{5}-4\sqrt{5}-12\sqrt{5}=-15\sqrt{5}\\ b,=2\sqrt{3}-\dfrac{2+\sqrt{3}}{1}=2\sqrt{3}-2-\sqrt{3}=\sqrt{3}-2\\ c,=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}}\\ =\dfrac{2\left(x+1\right)}{2\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}\)
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tính giá trị của đa thức q(x=(x2011+3.x2010-1)2012 khi x+3=0
Ta có:
\(Q\left(x\right)=\left[x^{1010}\left(x+3\right)-1\right]^{2012}=\left[x^{1010}.0-1\right]^{2012}=\left(-1\right)^{2012}=1\)
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Thực hiện phép tính:
a) \({x^2}.{x^4}\); b) \(3{x^2}.{x^3}\); c) \(a{x^m}.b{x^n}\) (a ≠ 0; b ≠ 0; m, n \(\in\) N).
a) \({x^2}.{x^4} = {x^{2 + 4}} = {x^6}\).
b) \(3{x^2}.{x^3} = 3.1.{x^{2 + 3}} = 3{x^5}\).
c) \(a{x^m}.b{x^n} = a.b.{x^{m + n}}\) (a ≠ 0; b ≠ 0; m, n \(\in\) N).
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Tính:
a) \(\sqrt{8x^3}\cdot\sqrt{2x}\left(x>0\right)\)
b) \(\sqrt{12x^5}\cdot\sqrt{3x}\left(x>0\right)\)
a) \(\sqrt{8x^3}\cdot2x\)
\(=\sqrt{8x^3\cdot2x}\)
\(=\sqrt{16x^4}\)
\(=\sqrt{\left(4x^2\right)^2}\)
\(=4x^2\)
b) \(\sqrt{12x^5}\cdot\sqrt{3x}\)
\(=\sqrt{12x^5\cdot3x}\)
\(=\sqrt{36x^6}\)
\(=\sqrt{\left(6x^3\right)^2}\)
\(=\left|6x^3\right|\)
\(=6x^3\)
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Cho hàm số: f(x)a(x+1)3+bxexf(x)a(x+1)3+bxex. Tìm a, b biết: f′(0)22f′(0)22 và ∫01f(x)dx5∫01f(x)dx5 m) ∫pi6pi4cos2xsin3xsin(x+pi4)dx∫pi6pi4cos2xsin3xsin(x+pi4)dxn) ∫0π2x−−√sinx−−√dx∫0π2xsinxdxp) ∫12dxx(x2012+1)dx∫12dxx(x2012+1)dxq) ∫03ln2dx(ex√3+2)2∫03ln2dx(ex3+2)2r) ∫1eln2x+lnx(lnx+x+1)3dx∫1eln2x+lnx(lnx+x+1)3dxs) ∫ln2ln3e2xex−1+ex−2√dx∫ln2ln3e2xex−1+ex−2dxt) ∫0pi3x+sin2x1+cos2xdx∫0pi3x+sin2x1+cos2xdxu)∫032x2+x−1x+1√dx∫032x2+x−1x+1dxv) ∫01x2ln(1+x2)dxAi nhanh mk tik nha.
Đọc tiếp
Cho hàm số: f(x)=a(x+1)3+bxexf(x)=a(x+1)3+bxex. Tìm a, b biết: f′(0)=22f′(0)=22 và ∫01f(x)dx=5∫01f(x)dx=5
m) ∫pi6pi4cos2xsin3xsin(x+pi4)dx∫pi6pi4cos2xsin3xsin(x+pi4)dx
n) ∫0π2x−−√sinx−−√dx∫0π2xsinxdx
p) ∫12dxx(x2012+1)dx∫12dxx(x2012+1)dx
q) ∫03ln2dx(ex√3+2)2∫03ln2dx(ex3+2)2
r) ∫1eln2x+lnx(lnx+x+1)3dx∫1eln2x+lnx(lnx+x+1)3dx
s) ∫ln2ln3e2xex−1+ex−2√dx∫ln2ln3e2xex−1+ex−2dx
t) ∫0pi3x+sin2x1+cos2xdx∫0pi3x+sin2x1+cos2xdx
u)∫032x2+x−1x+1√dx∫032x2+x−1x+1dx
v) ∫01x2ln(1+x2)dx
Ai nhanh mk tik nha.
a.Thực hiện phép tính:
A = \(-3\sqrt{8}+\sqrt{50}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
b.Rút gọn biểu thức
B = \(\left(\dfrac{5\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-x}\right).\left(1-\dfrac{1}{\sqrt{x}}\right)\) với x > 0 và x≠ 1
\(a,A=-3\sqrt{8}+\sqrt{50}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=-6\sqrt{2}+5\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=-\sqrt{2}-1+\sqrt{2}\)
\(=-1\)
Vậy \(A=-1\)
\(b,\)
\(=\left(\dfrac{5\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{5x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\dfrac{5\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{5\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{5\sqrt{x}-1}{\sqrt{x}}\left(đk:x>0,x\ne1\right)\)
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