Cho A =\(\frac{x^3+2x^2+3x+x^2\sqrt{4-x^2}+6}{\sqrt{x+3}+3}:\frac{x^2\left(\sqrt{x+2}+\sqrt{2-x}\right)+3\sqrt{x+2}}{2\sqrt{x+3}-3\sqrt{x+2}-\sqrt{x^2+5x+6}}\)
Rút gọn a
Những câu hỏi liên quan
Cho \(A=\frac{x^3+2x^2+3x+x^2\sqrt{4-x^2}+6}{\sqrt{x+3}+3}:\frac{x^2\left(\sqrt{x+2}+\sqrt{2-x}\right)+3\sqrt{x+2}}{2\sqrt{x+3}-3\sqrt{x+2}-\sqrt{x^2+5x+6}+6}\)
( ĐKXĐ: \(-2\le x< 2\))
a, Rút gọn A
b, Tìm Max của A
Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
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\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
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\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
Đặt \(\hept{\begin{cases}\sqrt{2+x}=a\\\sqrt{2-x}=b\end{cases}\Rightarrow}a^2+b^2=4\)
\(\Rightarrow C=\frac{\sqrt{2ab}.\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2ab}.\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}\)
\(=\sqrt{2ab}.\left(a-b\right)=\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{2+x}-\sqrt{2-x}\right)\)
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Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
Rút gọn:
\(A=\left(\frac{4x\sqrt{x}+3x+9}{x+5\sqrt{x}+6}-\frac{3-\sqrt{x}}{2+\sqrt{x}}\right)\div\left(\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{3+4\sqrt{x}}{x+5\sqrt{x}+6}\right)\)
\(B=\left(x-\sqrt{x}-2\right)\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)
A=\(\frac{6x-\left(x+6\right)\sqrt{x}-3}{2\left(x-4\sqrt{x}+3\right)\left(2-\sqrt{x}\right)}-\frac{3}{-2x+10\sqrt{x}-12}-\frac{1}{3\sqrt{x}-x-2}\)
Rút gọn
Rút gọn:
a/ \(\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+1}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
b/\(\frac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
Chỉ mìnhcách viết phân số đi, mình làm giùm cho.
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Câu 1 :A left(frac{x}{xsqrt{x}-4sqrt{x}}-frac{6}{3sqrt{x}-6}+frac{1}{sqrt{x}+2}right):left(sqrt{x}-2+frac{10-x}{x+2}right)a, rút gọn Ab, Tìm X sao cho A2Câu 2 Afrac{15sqrt{x}-11}{x+2sqrt{x}-3}+frac{3sqrt{x}-2}{1-sqrt{x}}-frac{2sqrt{x}+3}{sqrt{x}+3}a, rút gọn A left(vớiright)xge0,xne1b, chúng minh rằng Alefrac{2}{3}Câu 3 left(frac{2}{sqrt{x}}+frac{sqrt{x}}{sqrt{x}+2}right):frac{2sqrt{x}}{x+2sqrt{x}}left(vớix0right)a, Rút gọn Pb, tìm giá trị của x để P3
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Câu 1 :A= \(\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{x+2}\right)\)
a, rút gọn A
b, Tìm X sao cho A<2
Câu 2 \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a, rút gọn A \(\left(với\right)x\ge0,x\ne1\)
b, chúng minh rằng A\(\le\)\(\frac{2}{3}\)
Câu 3 \(\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\left(vớix>0\right)\)
a, Rút gọn P
b, tìm giá trị của x để P=3
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
Câu 2 :
\(ĐKXĐ:\hept{\begin{cases}x\ne\frac{4}{25}\\x\ne9\\x\ne1\end{cases}}\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{\left(2-5\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Mik quên mất cách chứng minh rồi :((
Chỉ biết : Dấu " = " xảy ra : \(\Leftrightarrow x=0\)
\(\)