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a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
2: \(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\)
=>x^2-3x-4+x^2+3x-4=2x^2-2
=>2x^2-8=2x^2-2(loại)
3: \(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)+x^2\left(x+3\right)=-7x^2+3x\)
=>x^3-3x^2-x^2+3x+x^3+3x^2+7x^2-3x=0
=>2x^3+6x^2=0
=>2x^2(x+3)=0
=>x=0(nhận) hoặc x=-3(loại)
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ x+2009=0
⇔ x=-2009
vậy x=-2009 là nghiệm của pt
a) ( x2 + x )2 + 4( x2 + x ) = 12
<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0
<=> ( x2 + x + 2)2 - 16 = 0
<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0
<=> ( x2 + x + 6 )( x2 + x - 2) = 0
Do : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x
=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1) + 2( x - 1) = 0
<=> ( x - 1)( x + 2 ) = 0
<=> x = 1 hoặc : x = - 2
KL.....
b) Kuroba kaito làm rùi nhé 
Ai lm giúp mk vs câu nào cũng được. Ai làm xong sớm nhất sẽ được tick
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
a: \(=\dfrac{\left(2\cdot547+1\right)\cdot3}{547\cdot211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{2735}{547\cdot211}=\dfrac{5}{211}\)
b: x=7 nên x+1=8
\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)
\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-...-x^3-x^2+x^2+x-5\)
=x-5=7-5=2
Ta có:
\(x^2+\dfrac{1}{x^2}=7\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2-2=7\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\)
\(\Rightarrow x+\dfrac{1}{x}=3\) ( Vì x > 0 )
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3.3=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}=18\)
Ta lại có:\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=x^5+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^5}=x^5+\dfrac{1}{x^5}+18\)
Mặt khác:
\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=\left(x+\dfrac{1}{x}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)^2-2\right]\)
\(=\left(x+\dfrac{1}{x}\right)\left(7^2-2\right)\)
\(=3.47=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}+18=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=123\)
b. Ta có: \(x+\dfrac{1}{x}=4\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3x^2.\dfrac{1}{x}+3x.\dfrac{1}{x^2}=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+12=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}=52\)
Lại có: \(x+\dfrac{1}{x}=4\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=16\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=14\)
Ta có: \(\left(x^3+\dfrac{1}{x^3}\right)\left(x^2+\dfrac{1}{x^2}\right)=52.14\)
\(\Rightarrow x^5+x+\dfrac{1}{x}+\dfrac{1}{x^5}=728\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=724\)
a.
\(A=x^7+\dfrac{1}{x^7}\)
Ta có: \(\left(x^5+\dfrac{1}{x^5}\right)\left(x^2+\dfrac{1}{x^2}\right)=728.14\)
\(\Rightarrow x^7+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^7}=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}+52=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}=10140\)