Cho biểu thức I = \(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\).\(\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn I
a) Tính giá trị của I với a = 16, b = 4
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
c. rút gọn biểu thức
\(C=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}-\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\) vs \(a\ge0,a\ne4,a\ne9\)
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 )
a
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
Cho biểu thức:
\(D=\left(\dfrac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\dfrac{a+b+2ab}{1-ab}\right)\)
a) Tìm đkxđ và rút gọn \(D\)
b) Tính \(D\) với \(a=\dfrac{2}{2+\sqrt{3}}\)
c) Tìm giá trị lớn nhất của \(D\)
P=\(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}-\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
a) Rút gọn
b) Tính P khi a=16 và b=4
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
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=\(\dfrac{1}{4}\)
\(A=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(B=\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}-a}\dfrac{a+b}{\sqrt{ab}}\right)\)
a. Rút gọn biểu thức
b. Tìm giá trị nguyên của x để biểu thức có giá trị nguyên
chỗ đầu mình nhầm B = \(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(....\right)\)
Cho biểu thức \(M=\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}-\dfrac{a}{\sqrt{a}+\sqrt{b}}-\dfrac{b}{\sqrt{b}-\sqrt{a}}\) với a,b>0 và \(a\ne b\) . Rút gọn M và tính giá trị biểu thức M biết \(\left(1-a\right).\left(1-b\right)+2\sqrt{ab}=1\)
\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\\ M=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\\ \Leftrightarrow1-a-b+ab+2\sqrt{ab}=1\\ \Leftrightarrow a+b-ab-2\sqrt{ab}=0\\ \Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-\sqrt{b}=\sqrt{ab}\\\sqrt{a}-\sqrt{b}=-\sqrt{ab}\end{matrix}\right.\)
Với \(\sqrt{a}-\sqrt{b}=\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)
Với \(\sqrt{a}-\sqrt{b}=-\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{-\sqrt{ab}}=-1\)
\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\left(\sqrt{a}-\sqrt{b}\right)+b\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\)
\(\Leftrightarrow a+b-ab-2\sqrt{ab}=0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\Leftrightarrow\sqrt{a}-\sqrt{b}=\sqrt{ab}\)
\(M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)
Rút gọn các biểu thức sau :
a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) - \(\sqrt{54}\)
b) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\) - a\(\sqrt{\dfrac{1}{a}}\)
c) ( \(\sqrt{28}\) - 2\(\sqrt{3}\) +7) \(\sqrt{7}\) +\(\sqrt{84}\)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
