CMR với \(a\ge0\)
\(\sqrt{a+\sqrt{a+\sqrt{a+...+\sqrt{a}}}}< \sqrt{a}+1\)\(\left(n\in N;n\ge1\right)\)
có n dấu căn
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Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
a) Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{1+\sqrt{a}}\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{1+\sqrt{a}}\)
\(=1+\sqrt{a}\) Bằng 1 kiểu gì đây._.?
a) Xin lỗi sửa lại phần a:
Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
b) Ta có: \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{6}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\)
1: ĐKXĐ: a>=0; a<>16
\(A=\dfrac{-2a-\sqrt{a}+a-16+a+4\sqrt{a}+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}+4\right)}\)
\(=\dfrac{3\sqrt{a}-12}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-4\right)}=\dfrac{3}{\sqrt{a}+2}\)
2: \(C-\dfrac{3}{2}=\dfrac{3}{\sqrt{a}+2}-\dfrac{3}{2}=\dfrac{6-3\sqrt{a}-6}{2\left(\sqrt{a}+2\right)}=\dfrac{-3\sqrt{a}}{2\left(\sqrt{a}+2\right)}< =0\)
=>C<=3/2
=>0<=C<=3/2
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)
Đề của bạn bị sai, mình sửa lại đề ở dưới nhé!
\(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(a+1\right)\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(a+1\right)^2-\left(\sqrt{a}\right)^2}\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{\left(a-1\right)^2-a}\)
\(=\frac{a^3-a^2\sqrt{x}+a^2-a\sqrt{a}+a-\sqrt{a}-a^3-a^2\sqrt{a}-a^2-a\sqrt{a}-a-\sqrt{a}+\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{a^2+2a+1-a}\)
\(=\frac{-2a^2\sqrt{a}-2a\sqrt{a}-2\sqrt{a}+\left(a+1\right)\left(a^2+2a+1-a\right)}{a^2+a+1}\)
\(=\frac{-2\sqrt{a}\left(a^2+a+1\right)+\left(a+1\right)\left(a^2+a+1\right)}{a^2+a+1}\)
\(=\frac{\left(a^2+a+1\right)\left[-2\sqrt{x}+\left(x+1\right)\right]}{a^2+a+1}\)
\(=x-1-2\sqrt{x}\)
\(=\left(\sqrt{x}-1\right)^2\)
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