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18 tháng 7 2017
\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
Vậy \(A< B\)
BM
1
2 tháng 12 2019
a) \(Q=\frac{x^4-x^2+2x+2}{x^4+x^3+x+1}\)
\(Q=\frac{x^2\left(x^2-1\right)+2\left(x+1\right)}{x^3\left(x+1\right)+\left(x+1\right)}\)
\(Q=\frac{x^2\left(x+1\right)\left(x-1\right)+2\left(x+1\right)}{\left(x+1\right)\left(x^3+1\right)}\)
\(Q=\frac{\left(x+1\right)\left[x^2\left(x-1\right)+2\right]}{\left(x+1\right)\left(x^3+1\right)}\)
\(Q=\frac{x^3-x^2+2}{x^3+1}\)
b) \(Q=\left|Q\right|=\frac{x^3-x^2+2}{x^3+1}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
28 tháng 6 2017
a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)
B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)
=\(\frac{3-x^2}{\left(x-1\right)^2}\)
b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)
TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)
c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2
30 tháng 11 2018
x khác 1
\(N=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2+4}{\left(x+1\right)\left(x^2+x+1\right)}\)
\(N=\frac{x^2+2x-x-2-2x^2-2x-2+2x^2+4}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{x^2+x+1}\)
Xét hiệu 1/3-N=\(\frac{1}{3}-\frac{x}{x^2+x+1}=\frac{x^2+x+1-3x}{3\left(x^2+x+1\right)}=\frac{x^2-2x+1}{3\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{3\left(x^2+x+1\right)}>0\)với mọi x khác 1
=> 1/3 >N
NH
1
25 tháng 7 2017
y=\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^8-1\right)\left(2^8+1\right)\)
=>y=\(2^{16}-1\)<\(2^{16}\)=x
=>x>y.
Vậy x>y
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\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\)
\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\)(*)
Đặt \(x^2-5x+4=a\)
(*)\(\Leftrightarrow a\left(a+2\right)\)
\(=a^2+2a\)
\(=a^2+2a+1-1\)
\(=\left(a+1\right)^2-1\ge-1\forall a\)
Dấu "=" xảy ra \(\Leftrightarrow a=-1\Leftrightarrow x^2-5x+4=-1\)
\(\Leftrightarrow x^2-5x+5=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\left(\frac{\pm\sqrt{5}}{2}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+5}{2}\\x=\frac{-\sqrt{5}+5}{2}\end{cases}}\)