Cho biểu thức \(Q=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)
a) Tìm ĐKXĐ của Q
b) Rút gọn biểu thức Q
c) Tìm x để Q < 0
d) Cho P = x-2. Tìm giá trị nhỏ nhất của PQ
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30 tháng 5 2023
a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)
=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)
Dấu = xảy ra khi x=1
17 tháng 3 2021
TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)
21 tháng 7 2023
a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)
c: Q=3
=>3căn x+6=căn x-2
=>2căn x=-8(loại)
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
=>2căn x-4-căn x-2>0
=>căn x>6
=>x>36
d: Q nguyên
=>căn x+2-4 chia hết cho căn x+2
=>căn x+2 thuộc Ư(-4)
=>căn x+2 thuộc {2;4}
=>x=0 hoặc x=4(nhận)
T
29 tháng 12 2023
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
MN
7 tháng 3 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
14 tháng 12 2018
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)
b: \(Q=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)
\(=\left(\dfrac{\left(x+2\right)\left(x+1\right)+2\cdot3x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right)\cdot\dfrac{x+1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2\left(4x^2-1\right)}{3x\cdot2\cdot\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{3x\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2x+1}{3x}+\dfrac{x^2-3x-1}{3x}=\dfrac{x^2-x}{3x}=\dfrac{x-1}{3}\)
c: Để Q<0 thì \(\dfrac{x-1}{3}< 0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-1;\dfrac{1}{2}\right\}\end{matrix}\right.\)
d: \(P\cdot Q=\dfrac{\left(x-1\right)}{3}\cdot\left(x-2\right)=\dfrac{1}{3}\left(x^2-3x+2\right)\)
\(=\dfrac{1}{3}\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{12}>=-\dfrac{1}{12}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=3/2
`#3107.101107`
`a)`
ĐKXĐ của Q: \(\left\{{}\begin{matrix}3x\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
`b)`
\(Q=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right)\div\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2-8x^2}{3x}\cdot\dfrac{1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2\left(1-4x^2\right)}{3x\left(2-4x\right)}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2\left(1-2x\right)\left(1+2x\right)}{3x\cdot2\left(1-2x\right)}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{1+2x}{3x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{1+2x-\left(3x+1-x^2\right)}{3x}\)
\(=\dfrac{1+2x-3x-1+x^2}{3x}\)
\(=\dfrac{x^2-x}{3x}=\dfrac{x\left(x-1\right)}{3x}=\dfrac{x-1}{3}\)
`c)`
Để `Q < 0:`
\(\Rightarrow\dfrac{x-1}{3}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Theo ĐKXĐ: \(x\notin\left\{-1;0;\dfrac{1}{2}\right\}\)
`d)`
`P = x - 2`
\(\Rightarrow PQ=\left(x-2\right)\left(\dfrac{x-1}{3}\right)=\dfrac{1}{3}\left(x-2\right)\left(x-1\right)=\dfrac{1}{3}\left(x^2-3x+2\right)\)
\(=\dfrac{1}{3}\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\\ =\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{12}\)
`=> PQ = 1/3(x-3/2)^2 - 1/12 \ge - 1/12` (Thỏa mãn ĐKXĐ)
`=>` Dấu `"="` xảy ra khi: `(x - 3/2)^2 = 0 => x = 3/2.`