cho biểu thức A= (\(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\)) *\(\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
mọi người giúp em với, em cần gấp ạ, cảm ơn mọi người nhiều
Mọi người giúp mình vs ạ. Mình đang cần gấp
Bài 1:Cho biểu thức A=\(\left(\dfrac{2}{1+2x}+\dfrac{4x^2+1}{4x^2-1}-\dfrac{1}{1-2x}\right):\dfrac{2}{4x^2-1}\)
a) Rút gọn biểu thức
b)Tính giá trị của A tại x=0; x=-3; x=\(\dfrac{1}{2}\)
c) Tìm x để A=2
Bài 2: Tìm giá trị nguyên thử của n để biểu thức B=\(\dfrac{2n^2+5n-1}{2n-1}\)có giá trị nguyên
Cho \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
a) Rút gọn.
b) Tìm GTNN của P khi x>1
Em cần câu b ạ. Cảm ơn ạ.
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3-x^2-2x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3+x^2+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
Ai giúp em với em đang cần gấp :
Đây là bài rút gọn biểu thức:
A=\(\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}\)\(\left(x\ne\pm\sqrt{2}\right)\)
B=\(\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\left(x\ne-\sqrt{5}\right)\)
Khó quá em không nghĩ ra mọi người giúp em nhé
Em cảm ơn
\(A=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
rút gọn biểu thức sau A=\(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2-4x+4+4x\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x+1}{2x}\)
Rút gọn biểu thức sau: \(A=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
( x2−2x / 2x2+8 − 2x2 / 8−4x+2x2−x3 ).(1− 1/x − 2/x2 )
=[ x2−2x / 2(x2+4) − 2x2 / 2(x2+4)−x(x2+4) ]. x2−x−2 / x2
=[x2−2x / 2(x2+4) − 2x2 / (2−x)(x2+3)] . x2−x−2 / x2
=(x2−2x)(2−x)−4x2 / 2(2−x)(x2+4) . x2+x−2x−2 / x2
= −x(x2+4) / 2(2−x)(x2+4). (x+1)(x−2) / x2
=x+1 / 2x
a) Rút gọn biểu thức sau: \(A=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
cho biểu thức
M = \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right).\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
a) Rút gọn biểu thức M
b) Tính giá trị của M với x = \(\dfrac{1}{2}\)
a: \(M=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Thay x=1/2 vào M, ta được:
\(M=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)
Giúp mình với, mình đang cần gấp
\(\dfrac{x^3}{1-x}\) + x2 + x + 1
\(\dfrac{2x}{x^2+4x+4}\) + \(\dfrac{x+1}{x+2}\) + \(\dfrac{2-x}{\left(x+2\right)^2}\)
Xin cảm ơn <3
\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
Cho biểu thức:
\(P=\left(\dfrac{2x-3}{4x^2-12x+5}+\dfrac{3}{1-2x}-\dfrac{2x-8}{2x^2-13x+20}\right):\dfrac{21+2x-8x^2}{4x^2-3+4x}+1\)
a,Rút gọn P
b,Tính giá trị của P khi |x|=\(\dfrac{1}{2}\)
c,,Tìm giá trị nguyên của x để P nhận giá trị nguyên.
d,Tìm xđể P>0
Mọi người giúp với ạ!!!
ĐKXĐ: \(x\ne\left\{\dfrac{-3}{2};\dfrac{1}{2};\dfrac{7}{4};\dfrac{5}{2};4;\right\}\)
\(P=\left(\dfrac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(2x-5\right)\left(x-4\right)}\right)\div\dfrac{\left(7-4x\right)\left(2x+3\right)}{\left(2x-1\right)\left(2x+3\right)}+1\)
\(P=\left(\dfrac{2x-3-3\left(2x-5\right)-2\left(2x-1\right)}{\left(2x-1\right)\left(2x-5\right)}\right)\dfrac{2x-1}{7-4x}+1\)
\(P=\dfrac{-8x+14}{\left(2x-5\right)\left(7-4x\right)}+1=\dfrac{2}{2x-5}+1\)
b/ \(\left|x\right|=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Với \(x=\dfrac{1}{2}\Rightarrow P=\dfrac{2}{2.\dfrac{1}{2}-5}+1=\dfrac{1}{2}\)
Với \(x=\dfrac{-1}{2}\Rightarrow P=\dfrac{2}{2.\left(\dfrac{-1}{2}\right)-5}+1=\dfrac{2}{3}\)
c/ Để P nguyên \(\Rightarrow\dfrac{2}{2x-5}\) nguyên \(\Rightarrow2⋮\left(2x-5\right)\Rightarrow2x-5=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(2x-5=-2\Rightarrow x=\dfrac{3}{2}\left(l\right)\)
\(2x-5=-1\Rightarrow x=2\)
\(2x-5=1\Rightarrow x=3\)
\(2x-5=2\Rightarrow x=\dfrac{7}{2}\left(l\right)\)
Vậy \(x=\left\{2;3\right\}\) thì P nguyên
d/ \(P>0\Rightarrow\dfrac{2}{2x-5}+1>0\Rightarrow\dfrac{2x-3}{2x-5}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3>0\\2x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3< 0\\2x-5< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< \dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\x>\dfrac{5}{2}\end{matrix}\right.\)