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\(A=\dfrac{1-\sqrt{x}+x}{\sqrt{x}}\\ =\dfrac{\left(1-\sqrt{x}+x\right)\sqrt{x}}{x}\\ =\dfrac{\sqrt{x}-x+x\sqrt{x}}{x}\)
lm tiếp...
Áp dụng BĐT Cauchy ta có:
P = \(x^2+y^2+\dfrac{33}{xy}\) \(\ge\) \(\dfrac{\left(x+y\right)^2}{2}+\dfrac{33}{x+y}\) = \(\dfrac{4^2}{2}+\dfrac{33}{4}=\dfrac{65}{4}\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x+y=4\\x=y\end{matrix}\right.\) <=> \(x=y=2\)
Vậy ...............................
coi như giải hệ pt
\(\hept{\begin{cases}y=x+1\left(1\right)\\y^2-3y\sqrt{x}+2x=0\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow\left(y^2-3\sqrt{x}.y+\frac{9x}{4}\right)=\frac{9x}{4}-2x=\frac{x}{2}\\ \)
\(\left(y-\frac{3\sqrt{x}}{2}\right)^2=\left(\frac{\sqrt{x}}{2}\right)^2\Rightarrow\orbr{\begin{cases}y=\frac{3\sqrt{x}}{2}-\frac{\sqrt{x}}{2}=\sqrt{x}\\y=\frac{3\sqrt{x}}{2}+\frac{\sqrt{x}}{2}=2\sqrt{x}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=x+1\left(3\right)\\2\sqrt{x}=x+1\left(4\right)\end{cases}}\)
\(\left(3\right)\Leftrightarrow\orbr{\begin{cases}\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{1}{4}-1\left(vonghiem\right)\\\left(\sqrt{x}-1\right)^2=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\end{cases}}\)
Vậy chỉ có điểm x=1; y=2 thỏa mãn
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
check giùm mik
Lời giải:
ĐKXĐ: $x>0$
a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\)
b.
\(x=\frac{4}{4+2\sqrt{3}}=(\frac{2}{\sqrt{3}+1})^2\Rightarrow \sqrt{x}=\frac{2}{\sqrt{3}+1}\)
\(P=\frac{(\frac{2}{\sqrt{3}+1}+1)^2}{\frac{2}{\sqrt{3}+1}}=\frac{3+3\sqrt{3}}{2}\)
a: Ta có: \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
\(A=\dfrac{x^2+2x+17}{2\left(x+1\right)}=\dfrac{x^2-6x+9+8x+8}{2\left(x+1\right)}\\ =\dfrac{\left(x-3\right)^2+8\left(x+1\right)}{2\left(x+1\right)}=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\)
Vì \(\left(x-3\right)^2\ge0;2\left(x+1\right)>0\) (do \(x>-1\))
\(\Rightarrow\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}\ge0\Leftrightarrow A=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\ge4\)
Dấu "=" xảy ra khi \(x=3\)
Vậy....
bạn 8 rút gon cho 2 bằng 4 thì mất 2 còn x+1 thôi fk