ĐKXĐ : \(x\ne\pm3\)
a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{\left(x+1\right)\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3}{x+3}-\frac{x-1}{x+3}\right)\)
\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)
\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)
\(A=\frac{-2\left(4x+1\right)\left(x+3\right)}{\left(3-x\right)\left(3+x\right)4}\)
\(A=\frac{-\left(4x+1\right)}{2\left(3-x\right)}\)
\(A=\frac{4x+1}{2\left(x-3\right)}\)
b) \(\left|x-5\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)
Mà ĐKXĐ x khác 3 => ta xét x = 7
\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)
c) Để A nguyên thì 4x + 1 ⋮ 2x - 3
<=> 4x - 6 + 7 ⋮ 2x - 3
<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3
Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3
=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }
=> x thuộc { 2; 1; 5; -2 }
Vậy .....
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)
\(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)
\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)
\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)
b)
Có 2 trường hợp:
T.Hợp 1:
\(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)
thay vào A ta được: A=\(-\frac{13}{8}\)
T.Hợp 2:
\(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)
Vậy không tồn tại giá trị của A tại x=3
Vậy với x=7 thì A=-13/8
c)
\(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)
Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)
Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .
Vậy không có giá trị nguyên nào của x để A nguyên
Câu 1:
\(P=\sqrt{a\left(a+b+c\right)+bc}+\sqrt{b\left(a+b+c\right)+ac}+\sqrt{c\left(a+b+c\right)+ab}\)
\(P=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\)
Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)
\(P\le\frac{a+b+a+c}{2}+\frac{b+a+b+c}{2}+\frac{c+a+c+b}{2}\)
\(=\frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}\)
\(=\frac{\left(2a+a+a\right)+\left(2b+b+b\right)+\left(2c+c+c\right)}{2}\)
\(=\frac{4\cdot\left(a+b+c\right)}{2}=\frac{4\cdot2}{2}=4\)
Vậy \(maxP=4\Leftrightarrow a=b=c=\frac{2}{3}\)
Câu 2:
Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)
\(\sqrt{3a\cdot\left(a+2b\right)}\le\frac{3a+a+2b}{2}=\frac{4a+2b}{2}=2a+b\)
\(\Rightarrow a\sqrt{3a\cdot\left(a+2b\right)}\le a\left(2a+b\right)=2a^2+ab\)
Chứng minh tương tự : \(b\sqrt{3b\left(b+2a\right)}\le2b^2+ab\)
Cộng theo vế 2 BĐT :
\(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\le2a^2+2b^2+2ab\)
\(=2\left(a^2+b^2\right)+2ab\le2\cdot2+a^2+b^2=4+2=6\)( vì \(2ab\le a^2+b^2\))
Dấu "=" a=b=1
Áp dụng BĐT Cô-si ta có :
\(\frac{x^2}{x^4+yz}\le\frac{x^2}{2\sqrt{x^4yz}}=\frac{x^2}{2x^2\sqrt{yz}}=\frac{1}{2\sqrt{yz}}=\frac{1}{4}\cdot\frac{2}{\sqrt{yz}}\le\frac{1}{4}\cdot\left(\frac{1}{y}+\frac{1}{z}\right)\)
Chứng minh tương tự \(\frac{y^2}{y^4+zx}\le\frac{1}{4}\cdot\left(\frac{1}{z}+\frac{1}{x}\right);\frac{z^2}{z^4+xy}\le\frac{1}{4}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế : \(VT=\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+zx}+\frac{z^2}{z^4+xy}\le\frac{1}{4}\cdot\left(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{2}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)(1)
\(GT\Leftrightarrow\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=3\)
Áp dụng BĐT quen thuộc \(a^2+b^2+c^2\ge ab+bc+ca\)ta có :
\(3=\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\ge\sqrt{\frac{x}{yz}}\cdot\sqrt{\frac{y}{zx}}+\sqrt{\frac{y}{zx}}\cdot\sqrt{\frac{z}{xy}}+\sqrt{\frac{x}{yz}}\cdot\sqrt{\frac{z}{xy}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Do đó \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)(2)
Từ (1) và (2) suy ra \(VT\le\frac{1}{2}\cdot3=\frac{3}{2}\)( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có \(x\sqrt{y+1}=\frac{2x}{\sqrt{6}}\sqrt{\frac{3}{2}\cdot\left(y+1\right)}\le\frac{2x}{\sqrt{6}}\cdot\frac{\frac{3}{2}+y+1}{2}=\frac{x}{\sqrt{6}}\cdot\left(\frac{5+2y}{2}\right)=\frac{5x+2xy}{2\sqrt{6}}\)
Chứng minh tương tự \(y\sqrt{x+1}\le\frac{5y+2xy}{2\sqrt{6}}\)
Do đó \(Q\le\frac{5\left(x+y\right)+4xy}{2\sqrt{6}}\le\frac{5+\left(x+y\right)^2}{2\sqrt{6}}=\frac{6}{2\sqrt{6}}=\frac{\sqrt{6}}{2}\)
Dấu "=" xảy ra \(x=y=\frac{1}{2}\)
Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{y+1}=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2=3\\Q=b\left(a^2-1\right)+a\left(b^2-1\right)\end{cases}}\)
Ta có \(Q=a^2b-b+ab^2-a=\left(a+b\right)\left(ab-1\right)\le\sqrt{2\left(a^2+b^2\right)}\cdot\left(\frac{a^2+b^2}{2}-1\right)\)
\(=\sqrt{2\cdot3}\cdot\left(\frac{3}{2}-1\right)=\frac{\sqrt{6}}{2}\)
Dấu "=" \(x=y=\frac{1}{2}\)
