cho A = x + 4
chứng minh : \(\sqrt{A}\)>\(A\)
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Những câu hỏi liên quan
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
1) \(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Ta có:
\(A\cdot\sqrt{x}=25\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=5^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=16\\\sqrt{x}=-6\text{(vô lý)}\end{matrix}\right.\)
c) Ta xét hiệu:
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-4\)
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-\dfrac{4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Với \(x>0\) thì \(\left(\sqrt{x}-1\right)>0\) và \(\sqrt{x}>0\)
\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
Nên A > 4 (đpcm)
3 tháng 8 2023
1: \(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
2: A*căn x=25
=>(căn x+1)^2=25
=>căn x+1=5
=>x=16
3: \(A-4=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>A>4
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
TH
1
4 tháng 7 2023
Ta thấy \(A-4=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}-4\)
\(=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\sqrt{x}>0\) nên \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\). ĐTXR \(\Leftrightarrow x=1\).
Như vậy \(A-4\ge0\) \(\Leftrightarrow A\ge4\)
(không phải là \(A>4\) như trong đề đâu nhé, dấu "=" vẫn có thể xảy ra nếu \(x=1\))
H
11 tháng 7 2023
\(A+B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)
11 tháng 7 2023
A+B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
29 tháng 10 2021
a: \(A=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)