Giúp mình giải câu a với
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LH
0
MK
1
26 tháng 10 2023
a: \(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)+1=x-\sqrt{x}+1\)
b:
\(\dfrac{x}{12}=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
\(\Leftrightarrow x\cdot\dfrac{1}{12}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}\)
\(\Leftrightarrow\dfrac{x}{12}=\dfrac{1}{3}\)
=>x=36
Khi x=36 thì \(A=36-6+1=37-6=31\)
c: \(B=\dfrac{2\sqrt{x}}{A}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)
\(B-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}\)
\(=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
=>B<2
\(2\sqrt{x}>0;x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=>B>0
=>0<B<2
giải giúp mình câu 5 với a
12 tháng 5 2021
a) [\(\dfrac{-2}{3}\).(-3)]. (x.x^2).(y.y).z
=2x^3y^2z
- Bậc: 6
b) A=\(\dfrac{-1}{4}\)x^2y^2. \(\dfrac{4}{3}\)xy^3
= (\(\dfrac{-1}{4}\) . \(\dfrac{4}{3}\)). ( x^2.x). ( y^2.y^3)
= \(\dfrac{-1}{3}\)x^3y^5
- Bậc: 8
VV
0
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2021
\(\Leftrightarrow\sqrt{4x^2-4x+1}=3x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\4x^2-4x+1=\left(3x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\5x^2-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{5}\)
Giúp mình câu này với ! với giải thichs giúp mình nhe!^^
NT
0
a, B=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}}{x-1}\)
B = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}}{x-1}=\dfrac{x+\sqrt{x}-2\sqrt{x}}{x-1}\)
B=\(\dfrac{x-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}\left(x-1\right)}{x-1}=\sqrt{x}\)
C = \(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
C=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\) \(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có : \(B.C=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}.\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}-1=\dfrac{1}{3}\Leftrightarrow\sqrt{x}=\dfrac{1}{3}+1=\dfrac{4}{3}\)
\(\Leftrightarrow x=\dfrac{16}{9}\)