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\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)
\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)
a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)
\(=\dfrac{1}{x}\)
b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)
\(=\dfrac{3x+6}{x-2}\)
giải giúp mình bài này ới ạ mình đng cần gấp
Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
Câu 1: Sửa lạ đề chút nhé : 4x + 1 -> 4x -1
Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
=> \(\sqrt{2}.A\)= \(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)
= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)
= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)
Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)
nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2
=> \(A=2:\sqrt{2}=\sqrt{2}\)
Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)
=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)
=> \(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)
=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)
=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)
\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)