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2. \(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(Q=4x^2+5x-12x-15+2019\)
\(Q=4x^2-7x+2004\)
\(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\)
\(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)
\(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)
\(\Rightarrow Q\ge\frac{32255}{16}\)
\(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)
3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\) (do a+b=1)
\(T=4a^2-4ab+4a^2-6a^2-6b^2\)
\(T=-2a^2-4ab-2b^2\)
\(T=-2\left(a^2+2ab+b^2\right)\)
\(T=-2\left(a+b\right)^2\)
\(T=-2.1^2=-2.1=-2\) (do a+b=1)
\(a^3-a^2b+ab^2-6b^3=0\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(a^2b-ab^2\right)+\left(3ab^2-6b^3\right)=0\)
\(\Leftrightarrow a^2\left(a-2b\right)+ab\left(a-2b\right)+3b^2\left(a-2b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)
Vì \(a>b>0\Rightarrow a^2+ab+3b^2>0\)nên từ (1) ta có \(a-2b=0\Leftrightarrow a=2b\)
Giá trị biểu thức \(P=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)
ta có \(P=a^4+b^4+2-2-ab\)
AD BĐT cô si ta có
\(a^4+1\ge2a^2\) dấu = khi a=1
\(b^4+1\ge2b^2\) dấu = khi b =1
Khi đó \(P\ge2a^2+2b^2-2-ab\)
\(P\ge2\left(a^2+b^2+ab\right)-2-3ab\)
\(P\ge4-3ab\)( Thay \(a^2+b^2+ab=3\)vào ) (1)
mặt khác \(a^2+b^2\ge2ab\)
khi đó \(a^2+b^2+ab=3\ge2ab+ab=3ab\)
=> \(ab\le1\) (2)
từ (1) và (2)
ta có \(P\ge4-3ab\ge4-3=1\)
vậy P đạt GTNN là 1 khi a=b=1
1/ Thay x=-4 vao A -> A= \(\frac{-4}{-4+3}\)= 4
2/ B=\(\frac{2}{x-3}\)+\(\frac{x-15}{x^2-9}\)
B= \(\frac{2\left(x+3\right)+x-15}{\left(x-3\right)\left(x+3\right)}\)
B= \(\frac{2x+6+x-15}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3}{x+3}\)
c, B>A <=> \(\frac{3}{x+3}\)> \(\frac{x}{x+3}\)
<=> \(\frac{3}{x+3}\)- \(\frac{x}{x+3}\)> 0
<=> \(\frac{3-x}{x+3}\)>0
<=> 3-x <0 / >0 ( Đkxd x khác -3 )
x+3 <0 / >0
..............
...............................
Vậy ...
1) \(A=\frac{x}{x+3}\)( ĐKXĐ : \(x\ne-3\))
Với x = -4 ( tmđk ) thì giá trị của A là
\(A=\frac{-4}{-4+3}=\frac{-4}{-1}=4\)
2) \(B=\frac{2}{x-3}+\frac{x-15}{x^2-9}\)( ĐKXĐ : \(x\ne\pm3\))
\(B=\frac{2}{x-3}+\frac{x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{2x+6+x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{3x-9}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x+3}\)
3) Để B > A
=> \(\frac{3}{x+3}>\frac{x}{x+3}\)( ĐKXĐ : \(x\ne-3\))
<=> \(\frac{3}{x+3}-\frac{x}{x+3}>0\)
<=> \(\frac{3-x}{x+3}>0\)
Xét hai trường hợp :
1.\(\hept{\begin{cases}3-x>0\\x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-x>-3\\x>-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 3\\x>-3\end{cases}}\Leftrightarrow-3< x< 3\)( tmđk )
2. \(\hept{\begin{cases}3-x< 0\\x+3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-x< -3\\x< -3\end{cases}}\Leftrightarrow\hept{\begin{cases}x>3\\x< -3\end{cases}}\)( loại )
Vì x nguyên => x ∈ { -2 ; -1 ; 0 ; 1 ; 2 ; 3 }
Vậy ...
Vì \(a+b=3\)
\(\Rightarrow\left(a+b\right)^2=9\)
\(\Leftrightarrow a^2+b^2+2ab=9\)
\(\Leftrightarrow a^2+b^2=7\)
Vì \(a+b=3\)
\(\Leftrightarrow\left(a+b\right)^3=27\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=27\)
\(\Leftrightarrow a^3+b^3=18\)