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\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}}{x-2\sqrt{x}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
a.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 1\left(x\ge0,x\ne4\right)\)
\(\Leftrightarrow\sqrt{x}-3< \sqrt{x}-2\)
\(\Leftrightarrow3>2\)
Vay \(B< 1\left(\forall x\ge0,x\ne4\right)\)
Lát mình giải 2 câu kia,di ăn com cái
b.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< \frac{3}{2}\)
\(\Leftrightarrow2\sqrt{x}-6< 3\sqrt{x}-6\)
\(\Leftrightarrow x>0\)
Vay \(B< \frac{3}{2}\left(\forall x>0,x\ne4\right)\)
c.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-3>x-3\sqrt{x}+2\)
\(\Leftrightarrow x-4\sqrt{x}+5< 0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+1< 0\) (vo ly)
Vay khong co gia tri nao cua x thoa man \(B>\sqrt{x}-1\)
Đề bài này be bét quá, xin phép sửa lại
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne\left\{1;4\right\}\end{cases}}\)
\(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
\(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{x-4\sqrt{x}+3-2x+3\sqrt{x}-2+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
b) Ta có: \(P< 1\)
\(\Leftrightarrow-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}< 0\)
Mà \(\sqrt{x}+1\ge1>0\left(\forall x\right)\)
\(\Rightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x< 1\\x>4\end{cases}}\)
\(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne4\end{cases}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-\sqrt{x}-2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\sqrt{x}-2}\)
b) Để P < 1
=> \(\frac{1}{\sqrt{x}-2}< 1\)
<=> \(\frac{1}{\sqrt{x}-2}-1< 0\)
<=> \(\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)
<=> \(\frac{1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
<=> \(\frac{3-\sqrt{x}}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}3-\sqrt{x}>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}>-3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 9\\x< 4\end{cases}}\Leftrightarrow x< 4\)
2. \(\hept{\begin{cases}3-\sqrt{x}< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}< -3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>9\\x>4\end{cases}}\Leftrightarrow x>9\)
Kết hợp với ĐK => Với \(\orbr{\begin{cases}x\in\left\{0;2;3\right\}\\x>9\end{cases}}\)thì thỏa mãn đề bài
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
Tacó \(\Delta\)=(-7)2-4x1x2=41>0 =>\(\sqrt{_{ }x1}\)=\(\dfrac{7+\sqrt{41}}{2}\)=>\(_{x1}\)=\(\dfrac{\left(7+\sqrt{41}\right)^2}{4}\)=\(\dfrac{45+7\sqrt{41}}{2}\) =>\(\sqrt{_{ }x2}\)=\(\dfrac{7-\sqrt{41}}{2}\)=>\(_{x_2}\)=\(\dfrac{\left(7-\sqrt{41^{ }}\right)^2}{4}\)=\(\dfrac{45-7\sqrt{41}}{2}\) so sánh với điều kiện X>_0
\(A=2x+\sqrt{4-2x^2}=\sqrt{2}.\sqrt{2x^2}+\sqrt{4-2x^2}\)
áp dụng BĐT bunhiacopxki,ta có:
\(A^2\le\left(2+1\right)\left(2x^2+4-2x^2\right)=3.4=12\)
\(\Leftrightarrow A\le\sqrt{12}\)
dấu = xảy ra khi \(\frac{\sqrt{2}}{\sqrt{2}x}=\frac{1}{\sqrt{4-2x^2}}\Leftrightarrow4-2x^2=x^2\Leftrightarrow x=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\)
vậy Amax = \(\sqrt{12}\)khi x=\(\frac{2}{\sqrt{3}}\)
2) \(a.\sqrt{\left(-15\right)^2}=\left|15\right|=15\)
\(b.\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|=2-\sqrt{5}\)
\(c.\sqrt{\left(3-\sqrt{7}\right)^2}=\left|3-\sqrt{7}\right|=3-\sqrt{7}\)
\(d.2\sqrt{a^2}=\left|2\sqrt{a}\right|=2\sqrt{a}\)
bai1
a)
√x=5=> x=25
b)
√(x+1)=2; x+1=4;x=3
c)
√(x^2)=4=> x=±4; x>0=>x=4
d)
2√(x+1)=4=>√(x+1)=2
x+1=4;x=3
Câu a :
Đặt \(A=x+\sqrt{2-x^2}\) .
A đạt MIN khi\(x+\sqrt{2-x^2}\)đạt MIN tương đương \(\sqrt{2-x^2}\) đạt MIN .
Do \(\sqrt{2-x^2}\ge0\Leftrightarrow x=\pm\sqrt{2}\)
Vậy MIN A sẽ là \(-\sqrt{2}\) khi \(x=-\sqrt{2}\)
Câu b :
Vì \(\left\{{}\begin{matrix}\sqrt{x-2}\ge0\\\sqrt{4-x}\ge0\end{matrix}\right.\) . Nên áp dụng BĐT \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :
\(\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
Vậy GTNN của biểu thức là \(\sqrt{2}\) . Dấu \("="\) xảy ra khi :
\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Wish you study well !!