Chứng minh biểu thức A = \(x^3+2x^2+x-3\sqrt{3}\) = \(-3\)tại \(x=\sqrt{3}-1\)
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6 tháng 9 2023
a: Khi x=121 thì \(A=\dfrac{121+3}{11+3}=\dfrac{124}{14}=\dfrac{62}{7}\)
b: \(B=\left(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{x+3\sqrt{x}-2-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
c: P=A:B
\(=\dfrac{x+3}{\sqrt{x}+3}:\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}\)
\(=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}-2=2\cdot2-2=2\)
Dấu = xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)
=>\(\sqrt{x}+1=2\)
=>x=1(nhận)
9 tháng 5 2022
1, Thay x = 16 vào ta được \(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)
2, \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{-x+6\sqrt{x}-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}+3}\)
Ta có đpcm
31 tháng 3 2021
a) Ta có: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
21 tháng 8 2023
a: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Khi x=25 thì \(A=\dfrac{5+2}{5+3}=\dfrac{7}{8}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}+2}+\dfrac{x+4}{4-x}\)
\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-x-4}{x-4}\)
\(=\dfrac{5\sqrt{x}-10}{x-4}=\dfrac{5}{\sqrt{x}+2}\)
c: \(A\cdot B=\dfrac{5}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{5}{\sqrt{x}+3}\)
Để A*B>1 thì \(\dfrac{5}{\sqrt{x}+3}-1>0\)
=>\(\dfrac{5-\sqrt{x}-3}{\sqrt{x}+3}>0\)
=>\(2-\sqrt{x}>0\)
=>căn x<2
=>0<=x<4
8 tháng 5 2022
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
28 tháng 10 2021
a) ĐKXĐ: \(x\ge0,x\ne9\)
\(B=\dfrac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
b) \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}=0\left(đk:x\ge0\right)\)\(\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
11 tháng 7 2021
Ta có: \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x-3-2x+8-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Ta có: \(A-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-1\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
hay A>1