Rút gọn biểu thức:
Rút gọn biểu thức:
\(ĐK:a>0;a\ne1\\ A=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ A=\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}+\dfrac{2a+2}{\sqrt{a}}\\ A=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1+2a+2}{\sqrt{a}}\\ A=\dfrac{2\left(a+\sqrt{a}+1\right)}{\sqrt{a}}=\dfrac{2\sqrt{a}\left(a+\sqrt{a}+1\right)}{a}\)
B= 1:(\(\dfrac{x+2}{x\sqrt{x}-1} + \dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\))
a) Rút gọn B
b) So sánh B với 3
a) ĐKXĐ: \(x>0,x\ne1\)
\(B=1:\dfrac{\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)-\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) \(B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)
Áp dụng BĐT Cauchy cho 2 só dương:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}.1}{\sqrt{x}}}=2\)
\(\Rightarrow B=1+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge1+2=3\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
Giúp mình với ạ
Câu a dấu con chuột che mất cái số r bạn
a) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{16}}{\sqrt{16}-3}=\dfrac{4}{4-3}=4\)
b) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}< \dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x}-3< 2\sqrt{x}\Leftrightarrow\sqrt{x}>-3\left(đúng\forall x\right)\)
ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow2x-3\sqrt{x}-2=7\)
\(\Leftrightarrow2x-3\sqrt{x}-9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\sqrt{x}=3\)(do \(2\sqrt{x}+3\ge3>0\))
\(\Leftrightarrow x=9\left(tm\right)\)
Câu b ạ
\(b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-12\cdot\dfrac{1}{3}\sqrt{x-2}=-1\\ \Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=-1\\ \Leftrightarrow-2\sqrt{x-2}=-1\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\Leftrightarrow x-2=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)
Câu c ạ
\(=\dfrac{\sin^240^0}{\cos^240^0}\cdot\cos^240^0-3+1-\sin^240^0\\ =\sin^240^0-3+1-\sin^240^0=-2\)
Câu b ạ
b) \(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-8\sqrt{3}\)
\(=2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}=1-8\sqrt{3}\)
\(b,=2-\sqrt{3}+\dfrac{2\left(\sqrt{3}-1\right)}{2}-6\cdot\dfrac{4}{\sqrt{3}}\\ =2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\\ =1-8\sqrt{3}\)
giải phương trình nghiệm nguyên không âm
x^2+x+1=Y^2
2x^2+y^2+2y=100
6x+21y+88xy=123
38x+117y=109
\(x^2+x+1=y^2\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+1=y^2\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-y^2=-1\\ \Leftrightarrow\left(x-y+\dfrac{1}{2}\right)\left(x+y+\dfrac{1}{2}\right)=-1=\left(-1\right)\cdot1\\ TH_1:\left\{{}\begin{matrix}x-y+\dfrac{1}{2}=-1\\x+y+\dfrac{1}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-\dfrac{3}{2}\\x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=1\end{matrix}\right.\)
\(TH_2:\left\{{}\begin{matrix}x-y+\dfrac{1}{2}=1\\x+y+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{1}{2}\\x+y=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{x-4}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)