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1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
Đặt \(D=\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\) (D >/ 0 với mọi 1/2 < x)
\(\Rightarrow D^2=2\sqrt{4x-1}-2\sqrt{4x^2-4x+1}=2\sqrt{4x-1}-2\left|2x-1\right|=2\sqrt{4x-1}-2\left(1-2x\right)=4x-2+2\sqrt{4x-1}\)
\(\Rightarrow D=\sqrt{D^2}=\sqrt{4x-2+2\sqrt{4x-1}}=\left|\sqrt{4x-1}+1\right|=\sqrt{4x-1}+1\)
\(=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}\)
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1\)
\(=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.\)
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.....)
\(\frac{\sqrt{x^2}+\sqrt{4-4x+x^2+1}}{2x-1}\)
\(=\frac{x+2-2\sqrt{x}+1}{2x-1}\)
\(=1+\frac{4-2\sqrt{x}}{2x-1}\)
em lớp 8 chỉ làm được thế thôi
ĐK: \(x\ge\frac{1}{4}\)
Ta có \(A^2=4x+2\sqrt{4x^2-\left(4x-1\right)}=4x+2\sqrt{\left(2x-1\right)^2}\)
Với \(x\ge\frac{1}{2},A=4x+2\left(2x-1\right)=8x-2\)
Do \(A\ge0\) nên \(A=\sqrt{8x-2}\)
Với \(\frac{1}{4}\le x< \frac{1}{2},A^2=4x+2\left(1-2x\right)=2\)
Do \(A\ge0\) nên \(A=\sqrt{2}\)
\(A=\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
\(\Rightarrow A^2=2x+\sqrt{4x-1}+2x-\sqrt{4x-1}+2\sqrt{\left(2x+\sqrt{4x-1}\right)\left(2x-\sqrt{4x-1}\right)}\)
\(=4x+2\sqrt{4x^2-4x+1}\)
\(=4x+2\sqrt{\left(2x-1\right)^2}\)
\(\Leftrightarrow\orbr{\begin{cases}A^2=4x+4x-1\\A^2=4x-4x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}A^2=-1\left(loai\right)\\A^2=1\end{cases}}\)
\(\Leftrightarrow A=1\)( vì A>0 )
Vậy bt A =1