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a: \(\dfrac{5x}{x^2+x-6}=\dfrac{5x}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{4x+12+x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{4}{x-2}+\dfrac{x-12}{x^2+x-6}\)
=>\(A=\dfrac{4}{x-2};B=\dfrac{x-12}{x^2+x-6}\)
b: \(\dfrac{5x+31}{x^2-3x-10}=\dfrac{5x+31}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x-15+2x+46}{\left(x-5\right)\left(x+2\right)}=\dfrac{3}{x+2}+\dfrac{2x+46}{\left(x-5\right)\left(x+2\right)}\)
=>\(A=\dfrac{3}{x+2};B=\dfrac{2x+46}{\left(x-5\right)\left(x+2\right)}\)
c: \(\dfrac{3x+5}{\left(x-1\right)^2}=\dfrac{3x-3+8}{\left(x-1\right)^2}=\dfrac{3}{x-1}+\dfrac{8}{\left(x-1\right)^2}\)
=>\(A=\dfrac{3}{x-1};B=\dfrac{8}{\left(x-1\right)^2}\)
Lời giải:
Ta có:
$x^2-3x+11=(x-\frac{3}{2})^2+\frac{35}{4}\geq \frac{35]{4}$
$\Rightarrow \frac{31}{x^2-3x+11}\leq 31:\frac{35}{4}=\frac{124}{35}$
$\Rightarrow \frac{31}{x^2-3x+11}+15\leq \frac{649}{35}$
Vậy gtln của biểu thức là $\frac{649}{35}$ khi $x=\frac{3}{2}$
A=2(x^2-2.5/4x+25/16)-50/16+7
A=2(x-√10/5)^2+31/8
Vì(x-√10/5)^2>=0 với mọi x
=>A>=31/8
Chọn B
\(2x^2-5x+7=2\left(x^2-\dfrac{5}{2}x+\dfrac{25}{16}\right)-\dfrac{25}{8}+7=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}+7\ge\dfrac{31}{8}\)
ĐTXR⇔\(x=\dfrac{5}{4}\)
Vậy minA =\(\dfrac{31}{8}\)khi x=\(\dfrac{5}{4}\)
Đáp án: \(B:\dfrac{31}{8}\)
Ta có: \(\dfrac{10x-5}{18}+\dfrac{x+3}{12}\ge\dfrac{7x+3}{6}-\dfrac{12-x}{9}\)
\(\Leftrightarrow\dfrac{2\left(10x-5\right)}{36}+\dfrac{3\left(x+3\right)}{36}\ge\dfrac{6\left(7x+3\right)}{36}-\dfrac{4\left(12-x\right)}{36}\)
\(\Leftrightarrow20x-10+3x+9\ge43x+9-48+4x\)
\(\Leftrightarrow23x-1-47x+39\ge0\)
\(\Leftrightarrow-24x+38\ge0\)
\(\Leftrightarrow-24x\ge-38\)
hay \(x\le\dfrac{19}{12}\)
Vậy: S={x|\(x\le\dfrac{19}{12}\)}
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
a) Ta có: \(P=\left(\dfrac{3}{x+1}+\dfrac{x-9}{x^2-1}+\dfrac{2}{1-x}\right):\dfrac{x-3}{x^2-1}\)
\(=\left(\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x-9}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x-3}{x^2-1}\)
\(=\dfrac{3x-3+x-9-2x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x-3}\)
\(=\dfrac{2x-14}{x-3}\)
b) Ta có: \(x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Thay x=-3 vào biểu thức \(P=\dfrac{2x-14}{x-3}\), ta được:
\(P=\dfrac{2\cdot\left(-3\right)-14}{-3-3}=\dfrac{-20}{-6}=\dfrac{10}{3}\)
Vậy: Khi \(x^2-9=0\) thì \(P=\dfrac{10}{3}\)
c) Để P nguyên thì \(2x-14⋮x-3\)
\(\Leftrightarrow2x-6-8⋮x-3\)
mà \(2x-6⋮x-3\)
nên \(-8⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(-8\right)\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(\Leftrightarrow x\in\left\{4;2;5;1;7;-1;11;-5\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;7;11;-5\right\}\)
Vậy: Để P nguyên thì \(x\in\left\{4;2;5;7;11;-5\right\}\)
