\(\dfrac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a-b}{\sqrt{a}-\sqrt{b}}\)
Những câu hỏi liên quan
Rút gọn biểu thức sau:
a) A= \(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
b) B=\(\left(\dfrac{2}{\sqrt{a}-\sqrt{b}}-\dfrac{2\sqrt{a}}{a\sqrt{a}+b\sqrt{b}}.\dfrac{a\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\right):4\sqrt{ab}\)
giúp mình với ạ, mk cần gấp lắm
rút gọn biểu thức
Adfrac{sqrt{a}-1}{asqrt{a}-a+sqrt{a}}:dfrac{1}{a^2+sqrt{a}} với a 0
Bdfrac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}}+dfrac{sqrt{a}-sqrt{b}}{2sqrt{ab}}left(dfrac{sqrt{b}}{a-sqrt{ab}}+dfrac{sqrt{b}}{a+sqrt{ab}}right) với a0 b0 và a khác b
Cdfrac{asqrt{b}+b}{a-b}.sqrt{dfrac{ab+b^2-2sqrt{ab^3}}{aleft(a+2sqrt{b}right)+b}}:dfrac{1}{sqrt{a}+sqrt{b}} với ab0
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rút gọn biểu thức
A=\(\dfrac{\sqrt{a}-1}{a\sqrt{a}-a+\sqrt{a}}:\dfrac{1}{a^2+\sqrt{a}}\) với a >0
B=\(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\) với a>0 b>0 và a khác b
C=\(\dfrac{a\sqrt{b}+b}{a-b}.\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\) với a>b>0
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
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thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\)
b) \(\dfrac{a-b}{\sqrt{a}+\sqrt{b}}-\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b: \(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
=0
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Thực hiện phép tính.
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right).a^2b^2.\sqrt{\dfrac{n}{m}}\)
Giải chi tiết ra hộ mình với ạ, mình cảm ơn ạ.
Rut gon phuong trinh \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^6}+\sqrt{b^6}}{a-b}\)
A \(\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
B\(\dfrac{\sqrt{ab}-2b}{\sqrt{a}-\sqrt{b}}\)
C \(\dfrac{2b}{\sqrt{a}-\sqrt{b}}\)
D\(\dfrac{\sqrt{ab}-2a}{\sqrt{a}-\sqrt{b}}\)
Giup minh chon dap an dung
Cho biểu thức A = \(\left(\dfrac{\sqrt{ab}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}+1\right):\left(\dfrac{\sqrt{ab}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{a}-\sqrt{b}}-1\right)\)
Cho \(\sqrt{ab}+1=4.\sqrt{b}\), tìm max của biểu thức A.
Đăt\(\sqrt{a}\)=x, \(\sqrt{b}\)=y (x,y>0)
=>xy+1=4y => 4y≥ \(2\sqrt{xy}\)=>\(2\sqrt{y}\)≥\(\sqrt{x}\)=> 4y≥x=> 4≥ \(\dfrac{x}{y}\)=> \(\dfrac{1}{4}\)≤\(\dfrac{y}{x}\)=>\(\dfrac{-1}{4}\)≥\(\dfrac{-y}{x}\)
Xét:A=(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{y-x}\)+1):(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{x-y}\)-1)
= \(\dfrac{-2y^2\left(x+1\right)}{\left(x-y\right)\left(x+y\right)}\).\(\dfrac{\left(x-y\right)\left(x+y\right)}{2xy\left(x+1\right)}\)
=> A= \(\dfrac{-y}{x}\)≤\(\dfrac{-1}{4}\)
Dấu "=" xảy ra <=> xy=1 và x=4y <=> x=2, y=\(\dfrac{1}{2}\) <=> a =4, b=\(\dfrac{1}{4}\)
Vậy Max A =\(\dfrac{-1}{4}\) <=> a=4, b=\(\dfrac{1}{4}\)
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\(A=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(B=\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}-a}\dfrac{a+b}{\sqrt{ab}}\right)\)
a. Rút gọn biểu thức
b. Tìm giá trị nguyên của x để biểu thức có giá trị nguyên
chỗ đầu mình nhầm B = \(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(....\right)\)
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Bài: Cho M=\(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) . ( \(\dfrac{b}{a-\sqrt{ab}}\) + \(\dfrac{\sqrt{b}}{a+\sqrt{ab}}\) )
a) Tìm đk của a và b để M xác định
b) C/m M > 0
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a< >b\end{matrix}\right.\)
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồia: ĐKXĐ:
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồi
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cho M= \(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+a\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
a) tìm điều kiện a và b để M xác định
b) c/m M>0
Sửa đề: \(M=\frac{\sqrt{a}+\sqrt{b}+1}{a+a\cdot\sqrt{b}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
a: ĐKXĐ: a>0; b>0
b: Ta có: \(M=\frac{\sqrt{a}+\sqrt{b}+1}{a+a\cdot\sqrt{b}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\frac{\sqrt{a}+\sqrt{b}+1}{a\left(\sqrt{b}+1\right)}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\frac{\sqrt{a}+\sqrt{b}+1}{a\left(\sqrt{b}+1\right)}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\)
\(=\frac{\sqrt{a}+\sqrt{b}+1}{a\left(\sqrt{b}+1\right)}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\frac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+1}{a\left(\sqrt{b}+1\right)}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\frac{2\sqrt{ab}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+1}{a\left(\sqrt{b}+1\right)}+\frac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\left(\sqrt{a}+\sqrt{b}+1\right)\left(\sqrt{a}+\sqrt{b}\right)+\sqrt{a}\left(\sqrt{b}+1\right)}{a\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+1\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2+\sqrt{a}+\sqrt{b}+\sqrt{ab}+\sqrt{a}}{a\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+1\right)}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2+\sqrt{ab}+\sqrt{b}+2\sqrt{a}}{a\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+1\right)}\)
=>M>0
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Rút gọn biểu thứca) dfrac{left(sqrt{a}-sqrt{b}right)^2+4sqrt{ab}}{left(sqrt{a+sqrt{b}}right)^2-4sqrt{ab}}.dfrac{a-b}{left(sqrt{a}-sqrt{b}right)^2} left(đkxđ:ane b;age0;bge0right)b) dfrac{a+b-2sqrt{ab}}{sqrt{a}-sqrt{b}}-dfrac{a-b}{left(sqrt{a}+sqrt{b}right)^2}left(đkxđ:ane b;age0;bge0right)HELP ME PLSSSSSSSSSS
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Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
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a
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
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