Rút gọn:
A=\(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\frac{2}{\sqrt{a}+\sqrt{b}}\)
B=\(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(a-b\right)\left(a\sqrt{a}+a\right)}\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)
\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)
Điều kiện : a, b\(\ge0\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Rút gọn biểu thức:
\(a,\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(b,\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
Các bn xem bài này mk làm đúng không
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
VT=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(a\sqrt{a}-a\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a+\sqrt{b}}}\right)\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\frac{a-b}{a-b}=1\Rightarrow\left(=VP\right)\)
b)\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\)
VT=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\sqrt{a}+\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
=\(a+\sqrt{ab}-\sqrt{ab}-b=a-b\Rightarrow\left(=VP\right)\)
Đây là đề chứng minh hả !
Phần a , b đúng r
Nhưng phần b chỗ \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\) = a - b
Dùng hằng đẳng thức thức 3 như vậy sẽ hay hơn !
Chúc bạn học tốt!
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
\(\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)\)
\(\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a^3}+3\right)\left(a-b\right)}\right)\)
Rút gọn 2 biểu thức trên?
Ai giúp mình với, tks nhiều
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
\(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(2a+2\sqrt{ab}+2b\right)}
\)
a. Rút gọn P
b. Tìm giá trị nguyên của a để giá trị P nguyên
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
\(\)Cho biểu thức
\(B=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\left(\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right)\)
a, Rút gọn B
b, Tính B khi a=16, b=4