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NN
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2022
\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x^2+4x-2x^2+2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{x-2}\)
AH
Akai Haruma
Giáo viên
2 tháng 3 2021
Lời giải:
ĐK: $x\geq 0; x\neq 1$
\(A=\left[\frac{(\sqrt{x}-1)(x+2\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}+2)}\right].\frac{\sqrt{x}-1}{(\sqrt{x}-1)(2\sqrt{x}+3)}\)
\(=\left(\frac{x+2\sqrt{x}+2}{\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}+3}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\frac{1}{2\sqrt{x}+3}=\frac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(2\sqrt{x}+3)}=\frac{\sqrt{x}+1}{2\sqrt{x}+3}\)
22 tháng 11 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
CM
29 tháng 8 2019
Ta có: P = x − x + 2 ( x + 1 ) ( x − 2 ) − x x ( x − 2 ) : 1 − x 2 − x = x − x + 2 − x ( x + 1 ) ( x + 1 ) ( x − 2 ) . 2 − x 1 − x = 2 − 2 x ( x + 1 ) ( x − 1 ) = 2 ( 1 − x ) ( x + 1 ) ( x − 1 ) = − 2 x + 1
H9
HT.Phong (9A5)
CTVHS
12 tháng 7 2023
\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
a) \(A=\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(A=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(A=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(A=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2\)
\(A=\left[\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\right]^2\)
\(A=\left(x-1\right)^2\)
\(A=x^2+2x+1\)
3 tháng 8 2023
\(A=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+2\sqrt{x}+3-2x+3\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) (ĐK: \(x\ne4;x\ne9;x\ge0\))
\(A=\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(A=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-9-\left(x-3\sqrt{x}+\sqrt{x}-3\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-9-x+2\sqrt{x}+3-2x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
22 tháng 11 2021
\(a,ĐK:x\ne\pm2\\ A=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\\ ĐK:x\ne-1;x\ne-2\\ B=\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ b,x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \forall x=0\Leftrightarrow A=\dfrac{1}{0-2}=-\dfrac{1}{2}\\ \forall x=-1\Leftrightarrow A=\dfrac{1}{-1-2}=-\dfrac{1}{3}\)
\(x^2+2x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ \Leftrightarrow B=\dfrac{1}{0+2}=\dfrac{1}{2}\)
12 tháng 1 2023
a: \(A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)