rút gọn biểu thức
A = \(\dfrac{x+2}{\left|x^2-1\right|}+\dfrac{x^2}{x+1}\)
B = \(2x:\dfrac{1}{2}x+\left(x+1\right)^2\)
K
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18 tháng 11 2021
\(ĐK:x\ne\pm1\)
Với \(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\Leftrightarrow A=\dfrac{x+2}{x^2-1}+\dfrac{x^2}{x+1}=\dfrac{x+2+x^3-x^2}{\left(x-1\right)\left(x+1\right)}\)
Với \(-1< x< 1\Leftrightarrow A=\dfrac{x+2}{1-x^2}+\dfrac{x^2}{x+1}=\dfrac{x+2+x^3-x^2}{\left(x+1\right)\left(1-x\right)}\)
\(B=2x\cdot\dfrac{2}{x}+x^2+2x+1\left(x\ne0\right)=x^2+2x+5\)
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
26 tháng 11 2021
\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
\(B=\dfrac{7a-7b+8a+8b-16b}{\left(a+b\right)\left(a-b\right)}=\dfrac{15a-15b}{\left(a-b\right)\left(a+b\right)}\\ B=\dfrac{15\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}=\dfrac{15}{a+b}\)
AH
Akai Haruma
Giáo viên
19 tháng 4 2021
Lời giải:
ĐK: $x\neq \pm 2; x\neq 0$
a)
\(A=\left[\frac{x+2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}\right].\frac{2-x}{x}=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}\)
\(=\frac{4x}{(x-2)(x+2)}.\frac{-(x-2)}{x}=\frac{-4}{x+2}\)
b) Để $A=1\Leftrightarrow \frac{-4}{x+2}=1$
$\Leftrightarrow x+2=-4$
$\Leftrightarrow x=-6$ (thỏa ĐKXĐ)
Vậy $x=-6$