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ĐK: x\(\ge0,x\ne1\)
a) \(Q=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Ta có \(Q=0,5\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=0,5\Leftrightarrow2-5\sqrt{x}=0,5\sqrt{x}+1,5\Leftrightarrow0,5=5,5\sqrt{x}\Leftrightarrow\sqrt{x}=\dfrac{1}{11}\Leftrightarrow x=\dfrac{1}{121}\left(tm\right)\)
Vậy \(x=\dfrac{1}{121}\) thì \(Q=0,5\)
c) Ta có \(Q=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}=\dfrac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)
Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}+\left(-5\right)\le\dfrac{2}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{2}{3}\Leftrightarrow Q\le\dfrac{2}{3}\)
Dấu bằng xảy ra khi x=0
Vậy GTLN của Q=\(\dfrac{2}{3}\)
Lời giải:
ĐKXĐ: \(x\geq 0, x\neq 1\)
Ta có:
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}+3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{x+2\sqrt{x}+3}-\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{3x+7\sqrt{x}-6}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{2x+\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}+3)}\)
\(=\frac{-5x+7\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(\sqrt{x}-1)(2-5\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b)
\(A=\frac{1}{2}\Leftrightarrow \frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow 2(2-5\sqrt{x})=\sqrt{x}+3\)
\(\Leftrightarrow 1=11\sqrt{x}\Rightarrow x=\frac{1}{121}\)
c)
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{17-5(\sqrt{x}+3)}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}-5\)
Ta thấy: \(\sqrt{x}\geq 0\Rightarrow \sqrt{x}+3\geq 3\Rightarrow A=\frac{17}{\sqrt{x}+3}-5\leq \frac{17}{3}-5=\frac{2}{3}\)
Vậy \(A_{\max}=\frac{2}{3}\)
Dấu bằng xảy ra khi $x=0$
Tất cả 3 bài này đều chung một dạng, bậc tử lớn hơn bậc mẫu nên đều không tồn tại GTLN mà chỉ tồn tại GTNN. Cách tìm thường là chia tử cho mẫu rồi khéo léo thêm bớt để sử dụng BĐT Cô-si
a) \(P=\dfrac{x+4}{4\sqrt{x}}=\dfrac{\sqrt{x}}{4}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}}{4}\dfrac{1}{\sqrt{x}}}=2.\dfrac{1}{2}=1\)
\(\Rightarrow P_{min}=1\) khi \(\dfrac{\sqrt{x}}{4}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=4\)
b) \(P=\dfrac{x+3}{2\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{2}+\dfrac{2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{2}+\dfrac{2}{\sqrt{x}+1}-1\)
\(\Rightarrow P\ge2\sqrt{\dfrac{\left(\sqrt{x}+1\right)}{2}\dfrac{2}{\left(\sqrt{x}+1\right)}}-1=2-1=1\)
\(\Rightarrow P_{min}=1\) khi \(\dfrac{\sqrt{x}+1}{2}=\dfrac{2}{\sqrt{x}+1}\Leftrightarrow x=1\)
c)ĐKXĐ: \(x\ge0\Rightarrow\) \(P=\dfrac{x-4}{\sqrt{x}+1}=\sqrt{x}-1-\dfrac{3}{\sqrt{x}+1}\)
\(P_{min}\) khi \(\dfrac{3}{\sqrt{x}+1}\) đạt max \(\Rightarrow\sqrt{x}+1\) đạt min, mà \(\sqrt{x}+1\ge1\) \(\forall x\ge0\) , dấu "=" xảy ra khi \(x=0\)
\(\Rightarrow P_{min}=-4\) khi \(x=0\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
b) A= \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\) + \(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)- \(\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
A= \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)}\) - \(\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)- \(\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}\)
= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c) GTLN (Max)
A= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
= -5+\(\dfrac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\)\(\ge\)0 (ĐKXĐ) \(\Rightarrow\) \(\sqrt{x}+3\ge3\)
\(\Rightarrow\) \(\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\)
\(\Rightarrow\) \(-5+\dfrac{17}{\sqrt{x}+3}\le-5+\dfrac{17}{3}\)
\(\Leftrightarrow\) A\(\le\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\sqrt{x}=0\) \(\Rightarrow\) \(x=0\)
Vậy Max A =\(\dfrac{2}{3}\) khi \(x=0\)
Bài 2:
a: \(\sqrt{4-x^2}>=0\)
Dấu '=' xảy ra khi x=2 hoặc x=-2
b: \(\sqrt{x^2-x+3}=\sqrt{x^2-x+\dfrac{1}{4}+\dfrac{11}{4}}\)
\(=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}>=\dfrac{\sqrt{11}}{2}\)
Dấu '=' xảy ra khi x=1/2
c: \(x+\sqrt{x}+1>=1\)
=>1/(x+căn x+1)<=1
Dấu '=' xảy ra khi x=0
GTNN và GTLN của cả A và B hay của A + B vậy bạn...
2.
\(P=\dfrac{\sqrt{x-2018}}{x+2}+\dfrac{\sqrt{x-2019}}{x}\)\(P=\dfrac{\sqrt{\left(x-2018\right).2020}}{\left(x+2\right)\sqrt{2020}}+\dfrac{\sqrt{\left(x-2019\right).2019}}{\sqrt{2019}.x}\)
Áp dụng BĐT AM-GM:
\(\sqrt{\left(x-2018\right).2020}\le\dfrac{1}{2}\left(x-2018+2020\right)=\dfrac{1}{2}\left(x+2\right)\)
\(\sqrt{\left(x-2019\right).2019}\le\dfrac{1}{2}\left(x-2019+2019\right)=\dfrac{1}{2}x\)
\(\Rightarrow P\le\dfrac{x+2}{2\sqrt{2020}\left(x+2\right)}+\dfrac{x}{2\sqrt{2019}.x}=\dfrac{1}{2\sqrt{2020}}+\dfrac{1}{2\sqrt{2019}}\)
\("="\Leftrightarrow x=4038\)
không phải bơ đâu, oan cho tớ quá :>
27/11 thi nên ít lên, với cả chị tớ cũng không cho chat :>
lấy mật khẩu của tớ vô đọc góc ib là biết mà :>
Lời giải:
ĐKXĐ: $x\geq 0$
Với $x\geq 0$ thì $-3\sqrt{x}\leq 0; \sqrt{x}+1>0$. Do đó: $A=\frac{-3\sqrt{x}}{\sqrt{x}+1}\leq 0$
Vậy $A_{\max}=0$. Giá trị này xác định tại $x=0$