Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
Q=a√a2−b2−(1+a√a2−b2):ba−√a2−b2=a√a2−b2−a2−(a2−b2)b√a2−b2=a√a2−b2−a2−a2+b2b√a2−b2=a−b√a2−
\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2+b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\frac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{b\left(a-b\right)}{b\sqrt{a^2-b^2}}=\frac{\left(a-b\right)}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)
:") Làm bừa nhezzz
a) \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2}-b^2}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(\sqrt{a^2-b^2}\right)^2}{b.\left(\sqrt{a^2-b^2}\right)}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{a^2-\left(a^2-b^2\right)}{b.\left(\sqrt{a^2-b^2}\right)}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
b) Thay a = 3b vào , ta được :
\(Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)
=(\(\frac{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}{\left(\sqrt{a+b}+\sqrt{a-b}\right)\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)+\(\frac{a-b}{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=(\(\frac{\sqrt{a^2-b^2}-\left(a-b\right)}{a+b-a+b}+\frac{\sqrt{a^2-b^2}+a-b}{a+b-a+b}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{2\sqrt{a^2-b^2}}{2b}\):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{\sqrt{a^2-b^2}}{b}\)*\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)
=\(\frac{a^2+b^2}{b}\)
b/ Thế \(b=a-1\)thì ta có
\(P=\frac{a^2+\left(a-1\right)^2}{a-1}=\frac{2a^2-2a+1}{a-1}\)
\(\Leftrightarrow2a^2-\left(2+P\right)a+1+P=0\)
\(\Rightarrow\Delta_a=\left(2+P\right)^2-4.2.\left(1+P\right)\ge0\)
\(\Leftrightarrow P\ge2+2\sqrt{2}\)
\(1a.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=21-2\sqrt{21}+2\sqrt{21}=21\) \(b.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)
\(2a.\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}.b^2}+\sqrt{\dfrac{a^2}{b^2}.\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+b\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}=\left(2+b\right)\sqrt{\dfrac{a}{b}}\) \(b.\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(x-1\right)^2}}.\sqrt{\dfrac{\left(2\sqrt{m}x-2\sqrt{m}\right)^2}{81}}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{\text{|}2\sqrt{m}x-2\sqrt{m}\text{|}}{9}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{2\sqrt{m}\text{|}x-1\text{|}}{9}=\dfrac{2m}{9}\) \(3a.VP=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1=VT\)
KL : Vậy đẳng thức được chứng minh.
\(b.VP=\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{\text{|}a+b\text{|}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{a+b}=\text{|}a\text{|}=VT\)
KL : Vậy đẳng thức được chứng minh .
P/s : Dài v ~
\(A=\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{x-2\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(2x-4\right)-4\sqrt{2x-4}+4}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}=\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
Đến đây có hai trường hợp :
- Với \(2\le x< 4\)\(\Rightarrow\left|\sqrt{2x-4}-2\right|=2-\sqrt{2x-4}\Rightarrow A=\frac{2-\sqrt{2x-4}}{2}\)
- Với \(x\ge4\Rightarrow\left|\sqrt{2x-4}-2\right|=\sqrt{2x-4}-2\Rightarrow A=\frac{\sqrt{2x-4}-2}{2}\)
b) \(B=\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}+a+1=a-\sqrt{a}-a-\sqrt{a}+a+1=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2\)
Rút gọn