Cho biểu thức C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
chứng minh C > 6
Những câu hỏi liên quan
Bleft(frac{xsqrt{x}+x+sqrt{x}}{xsqrt{x}-1}-frac{sqrt{x}+3}{1-sqrt{x}}right).frac{x-1}{2x+sqrt{x}-1} ĐKXĐ: ...frac{left(xsqrt{x}+x+sqrt{x}right)left(1-sqrt{x}right)-left(sqrt{x}+3right)left(xsqrt{x}-1right)}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2x+2sqrt{x}-sqrt{x}-1}frac{xsqrt{x}+x+sqrt{x}-x^2-xsqrt{x}-x-x^2+sqrt{x}-3xsqrt{x}+3}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2sqrt{x}left(sqrt{x}+1right)-left(sqrt{x}+1right)}frac{-3xsqrt{x}+2sqrt{x}-2x^2+3}{left(xsqrt{x}-1ri...
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\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Rút gọn và tìm GTNN của biểu thức:
\(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\left(\frac{x-1}{\sqrt{x}}\right)\left(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right)+\frac{2x+2}{\sqrt{x}}\)
\(A=2+\frac{2x+2}{\sqrt{x}}=\frac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)
\(A=2\left(\sqrt{x}+\frac{1}{\sqrt{x}}+1\right)\ge2\left(2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}+1\right)=6\)
Dấu "=" xảy ra khi \(x=1\) ko phù hợp ĐKXĐ nên \(A_{min}\) ko tồn tại
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ĐKXĐ:\(x\ne1;x>0\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\frac{x-1}{\sqrt{x}}.\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{x-1}\)
\(A=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+2+\frac{2}{\sqrt{x}}\ge2\sqrt{2\sqrt{x}.\frac{2}{\sqrt{x}}}+2=6\)
"="\(\Leftrightarrow x=1\)
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frac{x+1}{2left(x-1right)}+frac{2}{2left(sqrt{x}+1right)}+frac{sqrt{x}}{sqrt{x}left(sqrt{x}-1right)}.frac{left(x+1right).sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}-1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}+1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}.frac{xsqrt{x}+sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x-2sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x+2...
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\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
Tìm ĐKXĐ và rút gọn biểu thứcAfrac{left(sqrt{a}+sqrt{b}right)^2-4sqrt{ab}}{sqrt{a}-sqrt{b}}-frac{asqrt{b}+bsqrt{a}}{sqrt{ab}}Bleft(frac{2sqrt{x}-x}{xsqrt{x}-1}-frac{1}{sqrt{x}-1}right):frac{x-1}{x+sqrt{x}+1}Cleft(1-frac{x-3sqrt{x}}{x-9}right):left(frac{sqrt{x}-3}{2-sqrt{x}}+frac{sqrt{x}-2}{3+sqrt{x}}-frac{9-x}{x+sqrt{x}-6}right)Dleft(frac{sqrt{x}}{3+sqrt{x}}+frac{x+9}{9-x}right):left(frac{3sqrt{x}+1}{x-3sqrt{x}}-frac{1}{sqrt{x}}right)CM rằng GT của bthức A ko phụ thuộc vào aTìm x để C 4Tìm x sa...
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Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
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Cho biểu thức
\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{11}{x+\sqrt{x}+1}-\frac{34}{1-x\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\)
Cho biểu thức A= \(\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x-2}}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Rút gọn các biểu thức sau:
\(B=\frac{\sqrt{1+\sqrt{1-x^2}}[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}]}{x\left(2+\sqrt{1-x^2}\right)}\)
\(N=\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}\frac{1-x}{\sqrt{1-x^2}-1+x}\right).\left(\sqrt{\frac{1}{x^2}-1}-\frac{1-x}{x}\right).\frac{x}{1-x+\sqrt{1-x^2}}\)với -1<x<0
Mfrac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}}.left(frac{left(sqrt{x}-1right)left(x+sqrt{x}+1right)}{sqrt{x}left(sqrt{x}-1right)}right)-left(frac{left(sqrt{x}+1right)left(x-sqrt{x+1}right)}{sqrt{x}left(sqrt{x}+1right)}right)+frac{x+1}{sqrt{x}}left(frac{x+sqrt{x}+1}{sqrt{x}}-frac{x-sqrt{x}+1}{sqrt{x}}+frac{x+1}{sqrt{x}}right):sqrt{x}+1frac{x+2sqrt{x}+1}{sqrt{x}}:sqrt{x}+1frac{left(sqrt{x}+1right)^2}{sqrt{x}}.frac{1}{sqrt{x}+1}frac{sqrt{x}+1}{sqrt{x}}ĐÁP ÁN ĐÚNG KO???
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\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}.\)
=\(\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)-\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x+1}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\frac{x+1}{\sqrt{x}}\)
=\(\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\right):\sqrt{x}+1\)
=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}}:\sqrt{x}+1\)
=\(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}.\frac{1}{\sqrt{x}+1}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
ĐÁP ÁN ĐÚNG KO???
\(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
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Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
ĐKXĐ : x > 0 ; x khác 1
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
Rút gọn biểu thức:
a) Aleft(frac{3x-3sqrt{x}-3}{x+sqrt{x}-2}+frac{1}{sqrt{x}-1}-frac{1}{sqrt{x}+2}right):frac{1}{sqrt{x}+2}left(xge0,xne1right)
b) Bfrac{xsqrt{x}-3}{x-2sqrt{x}-3}-frac{2left(sqrt{x-3}right)}{sqrt{x}+1}+frac{sqrt{x}+3}{3-sqrt{x}}left(x0,xne9right)
c) Cfrac{2sqrt{x}-9}{x-5+6}-frac{sqrt{x}+3}{sqrt{x}-2}-frac{2sqrt{x}+1}{3-sqrt{x}}left(xge0,xne4,xne9right)
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Rút gọn biểu thức:
a) \(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\left(x\ge0,x\ne1\right)\)
b) \(B=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(x>0,x\ne9\right)\)
c) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)