Cho biểu thức:\(P=\left(\dfrac{2\text{x}}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3\text{x}}\right)v\text{ới}x\ne\pm3;x\ne0;x\ne5\)
1, Chứng minh \(P=\dfrac{x}{x-5}\)
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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 \(x\in\left\{4;2;5;7;11;-5\right\}\)
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
a) \(A=2x^2-\dfrac{1}{3}y\)
A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)
A=\(\dfrac{5}{3}\)\(x^2y\)
Tại \(x=2;y=9\) ta có
A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60
Vậy tại \(x=2;y=9\) biểu thức A= 60
b) P=\(2x^2+3xy+y^2\) (\(y^2\) là 1\(y^2\) nha bạn)
P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)
P= 6\(x^3y^3\)
Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có
P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)
Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)
c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)
=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)
=\(-\dfrac{1}{3}\)\(x^4y^2\)
Tại \(x=2;y=\dfrac{1}{4}\)ta có
\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)
\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)
CHÚC BẠN HỌC TỐT NHA
b) (4√x + 4)/(x + 2√x + 5) ≥ 1
⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
Do x ≥ 0 ⇒ x + 2√x + 5 > 0
⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0
⇔ 4√x + 4 - x - 2√x - 5 ≤ 0
⇔ -x + 2√x - 1 ≤ 0
⇔ -(x - 2√x + 1) ≤ 0
⇔ -(√x - 1)² ≤ 0 (luôn đúng)
Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0
a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+5}\)
b: 4(căn x+1)>=4
x+2căn x+5>=5
=>P<=4/5<1
\(B=\dfrac{\left(x^2-2x\right)\left(20x-11\right)}{\left(x-2012\right)\left(1982x^2+30\right)}-\dfrac{\left(20x-11\right)\left(x^2-3x+2012\right)}{\left(1982x^2+30\right)\left(x-2012\right)}\left(x\ne2012\right)\\ B=\dfrac{\left(20x-11\right)\left(x^2-2x-x^2+3x-2012\right)}{\left(x-2012\right)\left(1982x^2+30\right)} \\ B=\dfrac{\left(20x-11\right)\left(x-2012\right)}{\left(x-2012\right)\left(1982x^2+30\right)}=\dfrac{20x-11}{1982x^2+30}\)
1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)
\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)
\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)
\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)
\(=\dfrac{x}{x-5}\)