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đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4
Ta có: \(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)}{\sqrt{a}-2}+\frac{\sqrt{a}-2}{\sqrt{a}+2}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)^2+\left(\sqrt{a}-2\right)^2-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{-a-4\sqrt{a}-4+a-4\sqrt{a}+4-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{-4a-8\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=-\frac{4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=-\frac{4\sqrt{a}}{\sqrt{a}-2}\)
Ta có: \(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\)
\(=\frac{-2}{\sqrt{a}-2}+\frac{\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-2\sqrt{a}+\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
Ta có: \(\left(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\right):\left(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\right)\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}\cdot\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+3}=\frac{4a}{\sqrt{a}-3}\)
bài này cũng tương tự câu trên vậy tách màu ra là tính được mà . đâu có khó gì đâu bạn .
Biến đổi vế trái :vvv
\(VT=\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)
\(=\frac{a+b}{b^2}.\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)
\(=\frac{a+b}{b^2}.\frac{\left|ab^2\right|}{\left|a+b\right|}\)
\(=\frac{a+b}{b^2}.\frac{b^2.\left|a\right|}{a+b}=\left|a\right|=VP\left(đpcm\right)\)
( Vì a + b > 0 nên | a + b | = a + b ; b2 > 0 )
Đặt \(A=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{\left(a+\sqrt{a^2-b^2}\right)\left(a-\sqrt{a^2-b^2}\right)}{b\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)
\(A=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(A=\frac{a-b}{\sqrt{a-b}.\sqrt{a+b}}\)
\(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
Với \(a=3b\) ta có : \(A=\frac{\sqrt{a-b}}{\sqrt{a+b}}=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}}{2}\)
Chúc bạn học tốt ~
\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
\(\Rightarrow A=\sqrt{\left(2+x\right)^{^{ }3}}-\sqrt{\left(2-x\right)^3}=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)
\(\Rightarrow A=\dfrac{\sqrt{4+2\sqrt{4-x^2}}\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{\sqrt{2}\left(4+\sqrt{4-x^2}\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{2+x}+\sqrt{2-x}\right)\left(\sqrt{2+x}-\sqrt{2-x}\right)}{\sqrt{2}}=2\sqrt{2}\)
\(\sqrt{4\left(a-3\right)^2}\)
\(=\sqrt{4\left(a^2-6a+9\right)}\)
\(=\sqrt{4a^2-24a+36}\)
\(=\sqrt{\left(2a-6\right)^2}\)
\(=\left|2a-6\right|\)
\(=2a-6\)
\(\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}\)=( \(\sqrt{a}+\sqrt{b}\))( a + \(\sqrt{ab}\)+ b ) / \(\sqrt{a}+\sqrt{b}\)
= a + \(\sqrt{ab}\)+ b
\(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}=\sqrt{a}-\left(\sqrt{a}+2\right)=-2\)
Ta có: \(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}-\sqrt{a}-2=-2\)