Rút gọn:
A = \(\sqrt{\frac{x^2\left(x-1\right)^3}{1-x}}\)
Những câu hỏi liên quan
Bài 1:Rút gọna,frac{2}{sqrt{3}-1}-frac{2}{sqrt{3}+1}b,frac{2}{5-sqrt{3}}+frac{3}{sqrt{6}+sqrt{3}}c,left(1+frac{a+sqrt{a}}{1+sqrt{a}}right)timesleft(1-frac{a-sqrt{a}}{sqrt{a}-1}right)+aleft(ane1;age0right)Bài 2: Rút gọn biểu thứcPfrac{2sqrt{x}-9}{left(sqrt{x}-2right)left(sqrt{x}-3right)}-frac{sqrt{x}+3}{sqrt{x}-2}-frac{2sqrt{x}+1}{3-sqrt{x}}
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Bài 1:Rút gọn
\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(b,\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(c,\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)\times\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\left(a\ne1;a\ge0\right)\)
Bài 2: Rút gọn biểu thức
\(P=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
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rút gọn A=\(\frac{\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{2+\sqrt{1-x^2}}\)
ĐKXĐ: \(-1\le x\le1\)
Ta có: \(\sqrt{1+\sqrt{1-x^2}}=\frac{\sqrt{1+x+2\sqrt{1-x^2}+1-x}}{\sqrt{2}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{2}}\)
\(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)\)
\(\Rightarrow A=\frac{1}{\sqrt{2}}.\frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)}{2+\sqrt{1-x^2}}=\frac{1}{\sqrt{2x}}\)
1 \(P=\left(1-\frac{4}{\sqrt{x}+1}-\frac{1}{x-1}\right):\frac{x-2\sqrt{x}}{x-1}\)
rút gọn P
2 \(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
a, rút gọn A
b, tính P khi \(x=\frac{2}{2-\sqrt{3}}-2\sqrt{3}\)
cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
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Rút gọn: \(A=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
Rút gọn:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right]}{2-\sqrt{1-x^2}}\)
ChoQleft(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)left(frac{1-x}{sqrt{2}}right)^2a, rút gọnb, chứng minh nếu 0x1 thì Q0c, tìm GTLN của Q ChoAfrac{1}{2left(1+sqrt{x}+2right)}+frac{1}{2left(1-sqrt{x}+2right)}a, tìm x để a có nghĩab, rút gon Ac, tìm X nguyên để A nguyên ChoAleft(frac{sqrt{a}}{sqrt{a-1}}-frac{1}{a-sqrt{a}}right):left(frac{1}{sqrt{a}-1}-frac{2}{a-1}right)a, Rút gọn Ab, tính A Khi a3+2sqrt{2}
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\(ChoQ=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a, rút gọn
b, chứng minh nếu 0<x<1 thì Q>0
c, tìm GTLN của Q
\(ChoA=\frac{1}{2\left(1+\sqrt{x}+2\right)}+\frac{1}{2\left(1-\sqrt{x}+2\right)}\)
a, tìm x để a có nghĩa
b, rút gon A
c, tìm X nguyên để A nguyên
\(ChoA=\left(\frac{\sqrt{a}}{\sqrt{a-1}}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2}{a-1}\right)\)
a, Rút gọn A
b, tính A Khi a=3+\(2\sqrt{2}\)
\(\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}\left(\frac{\sqrt{x}+1}{3\sqrt{x}}-\sqrt{x}-1\right)\right]:\frac{\sqrt{x}-1}{\sqrt{x}}\)
a/ rút gọn
Bạn Ngọc Vĩ sai chỗ nhân phân phối nhé. Kq là \(\frac{2\sqrt{x}}{\sqrt{x}-1}\)
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ĐKXĐ: tự tìm nhoa
\(\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}.\frac{\sqrt{x}+1}{3\sqrt{x}}+2\left(\sqrt{x}-1\right)\right].\frac{\sqrt{x}}{\sqrt{x}-1}=\left(\frac{2}{3\sqrt{x}}-\frac{2}{3\sqrt{x}}+2\left(\sqrt{x}-1\right)\right).\frac{\sqrt{x}}{\sqrt{x}-1}=2\left(\sqrt{x}-1\right).\frac{\sqrt{x}}{\sqrt{x}-1}=2\sqrt{x}\)
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1. A left(sqrt{x}-frac{x+2}{sqrt{x}-1}right):left(frac{sqrt{x}}{sqrt{x}+1}-frac{sqrt{x}-4}{1-x}right)a. Rút gọn Ab. Tìm x để A0 c. Tìm giá trị nhỏ nhất A.2. Mleft(frac{2x+1}{sqrt{x^3}-1}-frac{1}{sqrt{x}-1}right):left(1+frac{x+4}{x+sqrt{x}+1}right)a. Rút gọn Mb. Tìm số nguyên x để M có giá trị nguyên 3. Nleft(frac{sqrt{a}+sqrt{b}}{1-sqrt{a.b}}+frac{sqrt{a}-sqrt{b}}{1+sqrt{a.b}}right):left(1+frac{a+b+2ab}{1-ab}right)a. Rút gọn Nb. Tính N khi afrac{2}{2-sqrt{3}}c. Tìm số nguyên a để N có giá trị ng...
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1. A= \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
a. Rút gọn A
b. Tìm x để A<0
c. Tìm giá trị nhỏ nhất A.
2. M=\(\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{x+4}{x+\sqrt{x}+1}\right)\)
a. Rút gọn M
b. Tìm số nguyên x để M có giá trị nguyên
3. N=\(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{a.b}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{a.b}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
a. Rút gọn N
b. Tính N khi a=\(\frac{2}{2-\sqrt{3}}\)
c. Tìm số nguyên a để N có giá trị nguyên
Gíup mình với. Cảm ơn nhiều ạ.
Rút gọn1.Aleft(frac{2sqrt{x}}{sqrt{x}+3}+frac{sqrt{x}}{sqrt{x}-3}-frac{3x+3}{x-9}right):left(frac{2sqrt{x}-2}{sqrt{x}-3}-1right)2.Bleft(frac{sqrt{a}+1}{sqrt{ab}+1}+frac{sqrt{ab}+sqrt{a}}{sqrt{ab}-1}-1right):left(frac{sqrt{a}+1}{sqrt{ab}+1}-frac{sqrt{ab}+sqrt{a}}{sqrt{ab}-1}+1right)3.Cleft(frac{2x-1+sqrt{x}}{1-x}+frac{2xsqrt{x}+x-sqrt{x}}{1+xsqrt{x}}right).left(frac{left(x-sqrt{x}right)left(1-sqrt{x}right)}{2sqrt{x}-1}right)
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Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
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 : \(\Leftrightarrow x=0\)
\(\)