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QA
29 tháng 7 2021
Trả lời:
a, \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-3}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\) \(\left(đkxđ:x\ge0;x\ne9\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-9}-\frac{2x-\sqrt{x}-3}{x-9}\)
\(=\frac{x-3\sqrt{x}}{x-9}+\frac{2x+3\sqrt{x}-9}{x-9}-\frac{2x-\sqrt{x}-3}{x-9}\)
\(=\frac{x-3\sqrt{x}+2x+3\sqrt{x}-9-2x+\sqrt{x}+3}{x-9}\)
\(=\frac{x+\sqrt{x}-6}{x-9}\)
CV
10 tháng 7 2018
1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
4 tháng 5 2021
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
AN
12 tháng 12 2019
a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
14 tháng 8 2020
ĐKXĐ: x \(\ge\)0; x \(\ne\)1 ; x \(\ne\)4
a) P = \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
P = \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}:\frac{\sqrt{x}\left(1-\sqrt{x}\right)-\sqrt{x}+4}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)
P = \(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\cdot\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-x-\sqrt{x}+4}\)
P = \(\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{4-x}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) P < 0 <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Do \(\sqrt{x}+2>0\) => \(\sqrt{x}-1< 0\) => \(\sqrt{x}< 1\) => \(x< 1\)
kết hợp với đk => S = {x| \(0\le x< 1\)}
c) P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Do \(\sqrt{x}+2\ge2\) => \(-\frac{3}{\sqrt{x}+2}\ge-\frac{3}{2}\) => \(1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra <=> x = 0
Vậy MinP = -1/2 khi x = 0
\(\sqrt{35-a\frac{23}{b}}\)
Đặt \(\sqrt{x}=a\Rightarrow A=\frac{a^2+9}{a+4}\Rightarrow A\cdot a+4A=a^2+9\Rightarrow a^2-A\cdot a+\left(9-4A\right)=0\)
Ta có:\(\Delta_a=A^2-4\left(9-4A\right)=A^2+16A-36\ge0\Leftrightarrow\left(A-2\right)\left(A+13\right)\ge0\)
hình như có gì đó sai sai bạn xem lại lời giải mình thử nha