Rút gọn biểu thức A = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
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Những câu hỏi liên quan
Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
NM
Cho Biểu thức M=\(y=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2_{-x}}\frac{ }{ }\)
Rút gọn biểu thức
2
T
5 tháng 8 2017
Điều kiện : \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{1}{1+\sqrt{x}}\right):\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{1+\sqrt{x}}\)
20 tháng 9 2019
A=(x+x+yy−xy):(xy+yx+xy−xy−xyx+y)
=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}=x+yx+xy+y−xy:xy(xy+y)(xy−x)x(xy−x)xy+y(xy+y)xy−(x+y)(xy+y)(xy−x)
=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}=x+yx+y:xy2−x2yx2y−x2xy+xy2+y2xy−y2xy+x2xy
=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}=x+yx+y.xy2+x2yxy2−x2y
=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}=x+yx+y.xy(x+y)xy(y−x)(x+y)
=\sqrt{y}-\sqrt{x}=y−x
NN
1
N
23 tháng 12 2018
ĐKXĐ: x khác 2 và -3
\(P=\frac{x+2}{x+3}-\frac{5}{x+2x-3x-6}-\frac{1}{x-2}\)
\(P=\frac{\left(x+2\right).\left(x-2\right)}{\left(x+3\right).\left(x-2\right)}-\frac{5}{\left(x-2\right).\left(x+3\right)}-\frac{x+3}{\left(x-2\right).\left(x+3\right)}\)
\(P=\frac{x^2-4-5-x-3}{\left(x-2\right).\left(x+3\right)}=\frac{x^2-x-12}{\left(x+2\right).\left(x+3\right)}=\frac{\left(x-4\right).\left(x+3\right)}{\left(x+2\right).\left(x+3\right)}=\frac{x-4}{x+2}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
T
28 tháng 7 2018
a) đk : \(x\ne2;-3\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{x^2+x-6}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{x^2+x-6}\)
\(=\frac{x^2-x-12}{x^2+x-6}\)
\(=\frac{x^2-4x+3x-12}{x^2+3x-2x-6}\)
\(=\frac{x\left(x-4\right)+3\left(x-4\right)}{x\left(x+3\right)-2\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b)
A>0.
\(\frac{x-4}{x-2}>0\)
th1 :
x-4>0 và x-2>0
<=> x>4
th2 : x-4 <0 và x-2 < 0
<=> x<2
Vậy để A>0 thì x>4 hoặc x<2
28 tháng 7 2018
a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\) \(\left(ĐKXĐ:x\ne2;-3\right)\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{-1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x^2-4x\right)+\left(3x-12\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x-4}{x-2}\)
b) Để \(A>0\)thì \(\frac{x-4}{x-2}>0\)
\(\Rightarrow\)(x - 4) ; (x - 2) cùng dấu
* hoặc \(\hept{\begin{cases}x-4>0\\x-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x>2\end{cases}}\Leftrightarrow x>4\)
* hoặc \(\hept{\begin{cases}x-4< 0\\x-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x< 2\end{cases}}\Leftrightarrow x< 2\)
Vậy \(\orbr{\begin{cases}x>4\\x< 2\end{cases}}\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
BT <=>
\(A=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-9-x-3}{MTC}=\frac{x^2-x-12}{MTC}\)
A = \(\frac{x+2}{x+3}\)\(-\frac{5}{X^2+X-6}\)\(+\frac{1}{2-X}\)
A= \(\frac{x+2}{x+3}\)\(-\frac{5}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{1}{X-2}\)
A = \(\frac{\left(X+2\right)\left(X-2\right)}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{5}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{X+3}{\left(X-2\right)\left(X+3\right)}\)
A= \(\frac{\left(X+2\right)\left(X-2\right)-5-\left(X+3\right)}{\left(X-2\right)\left(X+3\right)}\)
A= \(\frac{X-4-5-X-3}{\left(X-2\right)\left(X+3\right)}\)
A= \(-\frac{12}{\left(X-2\right)\left(X+3\right)}\)