C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)
Rút gọn các biểu thức
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0)
b) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0; a ≠ b)
c) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\) với a,b > 0
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{ab}\)
b) Giống câu a ?
c) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\)
\(=\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\sqrt{\frac{4b}{a}}+\sqrt{\frac{1}{ab}}\right):\left(\frac{ab+2b-a+1}{ab}\right)\)
\(=\frac{ab-a+2b+1}{\sqrt{ab}}\cdot\frac{ab}{ab+2b-a+1}\)
\(=\sqrt{ab}\)
Rút gọn biểu thức
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0; a ≠ b)
b) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)vớia,b>0\)
C/m biểu thức
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)=1\)(a,b>0,a\(\ne\)0
b)\(\frac{a-b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\left(a,b>0,a\ne b\right)\)
c)\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\left(a>0,a\ne1\right)\)
d)\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1-a\right)^2\left(a\ge0,a\ne1\right)\)
Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
Cho a>0, b>0, a≠b, rút gọn biểu thức:
A = \(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left[\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right].\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\left[\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(a-b\right)\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{a-b}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}\\ =\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right):\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\\ =\left(\frac{\sqrt{a^2}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{b^2}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right):\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}(\sqrt{a}-\sqrt{b})}\\ =\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}\)
Ngu vã đái 1s là ra ovs vật🤦♂️🤦♂️🤦♂️🤦♂️🤦♂️🤦♂️🤦♂️🦵👎👎👎👎👎👎👎👎👊👎👎😡😡😡👋👋👋👋👋👋✊✊✊✊✊
Rút gọn biểu thức : \(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{ab}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\) với a,b > 0 \(a\ne b\)
\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)
\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}+\frac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{a-b}+\frac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)
\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b+\sqrt{ab}+b+a-\sqrt{ab}}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)
\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{2\left(a+b\right)}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)
\(P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)
TH1: \(a>b\Rightarrow P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{a}-\sqrt{b}}{2}=0\)
TH2: \(0< a< b\Rightarrow P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{b}-\sqrt{a}}{2}=\sqrt{a}-\sqrt{b}\)
Rút gọn biểu thức : \(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{ab}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\) với a,b > 0 \(a\ne b\)
Cho biểu thức:
\(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\) với a > 0, b > 0, ab \(\ne\)1
a) Rút gọn D
b) Tính giá trị của D với a = \(\frac{2}{2-\sqrt{3}}\)
1) Rút gọn biểu thức sau :
M= \(\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{âb}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\)
với b>a>0
giúp mk với
Ch0 biểu thức \(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}-b\)với a>0, b>0.
a) Rút gọn A
b) Tìm b để A=1
10k vittel cho bạn nào nhanh nhất
a) ĐK: a > 0; b > 0
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}-b\)
\(=\frac{\sqrt{a}+\sqrt{b}+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}-b\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)-b\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}-b\)
\(=2\sqrt{b}-b\)
b) \(A=1\)\(\Rightarrow\)\(2\sqrt{b}-b=1\)
\(\Leftrightarrow\)\(b-2\sqrt{b}+1=0\)
\(\Leftrightarrow\) \(\left(\sqrt{b}-1\right)^2=0\)
\(\Leftrightarrow\)\(\sqrt{b}-1=0\)
\(\Leftrightarrow\)\(\sqrt{b}=1\)
\(\Leftrightarrow\)\(b=1\) (t/m ĐKXĐ)
Vậy b=1
