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theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
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)
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: \(\left(x^5+\dfrac{1}{x^5}\right)\left(x^2+\dfrac{1}{x^2}\right)=728.14\)
\(\Rightarrow x^7+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^7}=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}+52=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}=10140\)
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
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:
\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=\left(x+\dfrac{1}{x}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)^2-2\right]\)
\(=\left(x+\dfrac{1}{x}\right)\left(7^2-2\right)\)
\(=3.47=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}+18=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=123\)
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)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
(x-1/x):(x+1/x)=n
=>(x-1/x)=n*(x+1/x)
\(V=\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right):\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\)
\(=n\cdot\dfrac{\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)^2-2}\)