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\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
\(A=\left(a+2b-5+b\right)^2-2ab+34=\left(a+2b-5\right)^2+2b\left(a+2b-5\right)+b^2-2ab+34\)
\(A=\left(a+2b-5\right)^2+5b^2-10b+5+29\)
\(A=\left(a+2b-5\right)^2+5\left(b-1\right)^2+29\ge29\)
\(A_{min}=29\) khi \(\hept{\begin{cases}a=3\\b=1\end{cases}}\)
\(B=x+\frac{25}{x}-8\ge2\sqrt{x.\frac{25}{x}}-8=2\)
\(B_{min}=2\) khi \(x=5\)
\(C=\frac{x^2-15x+36}{x}=x+\frac{36}{x}-15\ge2\sqrt{x.\frac{36}{x}}-15=-3\)
\(C_{min}=-3\) khi \(x=6\)
Bđt phụ \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\forall\)
\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow a^2+b^2-2ab=\left(a-b\right)^2\ge0\)(đúng)
Áp dụng ta được :
\(A\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}=\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
Vậy \(A_{min}=\frac{25}{2}\) tại \(x=y=\frac{1}{2}\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
Áp dụng bđt: a2 + b2 > = (a + b)2/2
Cm đúng <=> 2a2 + 2b2 - a2 - 2ab - b2 > = 0
<=> (a - b)2 > = 0 (luôn đúng với mọi a,b
Khi đó, ta có: A = \(\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\ge\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
Áp dụng bđt: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
CM đúng <=> (a + b)2 > = 4ab
<=> (a - b)2 > = 0 (luôn đúng với mọi a,b)
Ta lại có: A \(\ge\frac{\left(2+\frac{4}{x+y}\right)^2}{2}=\frac{\left(2+\frac{4}{1}\right)^2}{2}=18\)
Dấu"=" xảy ra <=> x = y = 1/2
Vậy minA = 18/ <=> x = y = 1/2
a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)
Kết hợp với giả thiết vậy x = -1
Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)
c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )
d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)
\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)vì -2 < 0
e, tự làm nhéee
f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)
mà x + 4 > x + 2
\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)
Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )
Vậy ko có x nguyên để C < 0
g, Ta có : \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)
Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)
\(=\left(x+1\right)^2-9\ge-9\)
Dấu ''='' xảy ra khi x = -1
Vậy GTNN D là -9 khi x = -1
\(B=\frac{x^2+4x+85}{3\left(x+2\right)}=\frac{\left(x^2-14x+49\right)+\left(18x+36\right)}{3\left(x+2\right)}\)
\(=\frac{\left(x-7\right)^2+18\left(x+2\right)}{3\left(x+2\right)}=\frac{\left(x-7\right)^2}{3\left(x+2\right)}+6\ge6\forall x>0\)
Dấu "=" xảy ra khi: \(x-7=0\Leftrightarrow x=7\)