Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ĐS: .
b) ĐS: Nếu thì
Nếu ab
c) ĐS:
d)
Nhận xét. Nhận thấy rằng để có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.
a) ĐS: .
b) ĐS: Nếu thì
Nếu ab
c) ĐS:
d)
Nhận xét. Nhận thấy rằng để có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.
A = \(\dfrac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\) - \(\dfrac{a-b}{\sqrt{a+\sqrt{b}}}\)
= \(\dfrac{^{\left(\sqrt{a}+\sqrt{b}\right)^2}}{\sqrt{a}+\sqrt{b}}\) - \(\dfrac{a-b}{\sqrt{a+\sqrt{b}}}\)
= \(\left(\sqrt{a}+\sqrt{b}\right)\) - \(\dfrac{a-b}{\sqrt{a+\sqrt{b}}}\)
= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a-b\right)}{\sqrt{a}+\sqrt{b}}\)
=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
=\(2\sqrt{b}\)
đk : \(a\ge0;b\ge0;a\ne b\)
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{a+2\sqrt{ab}+b+a-2\sqrt{ab}+b}{a-b}\) = \(\dfrac{2\left(a+b\right)}{a-b}\)
b) đk : \(a\ge0;b\ge0;a\ne b\)
\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{1}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}\)
= \(\dfrac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)
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: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
Câu (A) đề có sao không nhỉ?
\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow\dfrac{1}{a-1}\)
\(E=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)
a) \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-32}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{8}.\sqrt{2x-1}+\sqrt{18}.\sqrt{2x-1}-\sqrt{32}.\sqrt{2x-1}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}.\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}.\sqrt{2}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}=\sqrt{5}\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow x=3\)
Vậy ...
b) \(\sqrt{x^2-6x+9}=x+3\)
\(\Leftrightarrow\sqrt{x^2-2.x.3+3^2}=x+3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=x+3\)
\(\Leftrightarrow\left|x-3\right|=x+3\)
\(\Leftrightarrow x-3=x+3\) hoặc \(x-3=-x-3\)
\(\Leftrightarrow x=0\)
Vậy ...
bài 2 :
A = \(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)
\(=\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a+\sqrt{b}}\right)}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{ab}-a-b}\right)\)
\(=\left(\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-a+\sqrt{ab}-b}\right)\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-\left(a-\sqrt{ab}+b\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\left(-\left(\sqrt{a}+\sqrt{b}\right)\right)\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right).\left(-1\right).\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=-\left(\sqrt{a}-\sqrt{b}\right)=\sqrt{b}-\sqrt{a}\)
cuối cùng cũng xong, mong bn phù hộ độ trì cho mk
bài 2 ) a) đk : \(a>0;b>0\)
b) P = \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
P = \(\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
P = \(\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) = \(a-b\)
c) ta có P = \(a-b\) thay \(a=2\sqrt{3};b=\sqrt{3}\) vào ta có
P = \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\) vậy khi \(a=2\sqrt{3};b=\sqrt{3}\) thì P = \(\sqrt{3}\)
bài 1) a) P = \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
P = \(\dfrac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-a^2\sqrt{a}+a^2-a+\sqrt{a}}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(\dfrac{2a^2-2a}{a^2-a}+\dfrac{2a+1}{\sqrt{a}}\) = \(\dfrac{2\left(a^2-a\right)}{a^2-a}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(2+\dfrac{2a+2}{\sqrt{a}}\) = \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b) ta có P = 7 \(\Leftrightarrow\) \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\) \(\Leftrightarrow\) \(2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow\) \(2a-5\sqrt{a}+2=0\) (1)
đặc \(\sqrt{a}=u\) \(\left(u\ge0\right)\) (1) \(\Leftrightarrow\) \(2u^2-5u+2\)
\(\Delta=\left(-5\right)^2-4.2.2\) = \(25-16=9>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(u_1=\dfrac{5+3}{4}=\dfrac{8}{4}=2\left(tmđk\right)\)
\(u_2=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\left(tmđk\right)\)
ta có : \(u=\sqrt{a}=2\Leftrightarrow x=4\)
\(u=\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(a=4;a=\dfrac{1}{4}\) thì P = 7
a)
\(\sqrt{\dfrac{27a^4}{48a^2}}=\sqrt{\dfrac{9a^2}{16}}=\sqrt{\left(\dfrac{3a}{4}\right)^2}=\dfrac{3a}{4}\)
b)
\(\dfrac{\sqrt{9x^2-25}}{\sqrt{3x+5}}=\dfrac{\sqrt{\left(3x-5\right)\left(3x+5\right)}}{\sqrt{3x+5}}=\sqrt{3x-5}\)
c)
\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\\ =\left(3-\sqrt{5}\right)+2.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\\ =2.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+6\\ =2.\sqrt{9-5}+6\\ =10\\ \Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\)
d) KO khó!!
\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}\)
\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}\)
\(B=a\sqrt{b}-b\sqrt{a}\)
Với `a,b > 0` có:
`B=[a\sqrt{b}-b\sqrt{a}]/\sqrt{ab} :1/[\sqrt{a}.\sqrt{b}]`
`B=[a\sqrt{b}-b\sqrt{a}]/[\sqrt{ab}] .\sqrt{ab}`
`B=a\sqrt{b}-b\sqrt{a}`