B=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4-2}}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4+2}}\)
Rút gọn
3.Rút gọn biểu thức :A=
\(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}}\)
mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)
ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)
<=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)
<=> \(A^3=x^3-3x+3A\)
<=> \(A^3-3A-x^3+3x=0\)
<=>\(\left(A^3-x^3\right)-3A+3x=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)
<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )
vậy \(A=x\)
rút gọn A= \(\frac{x^3+3x^2-4+\left(x^2-4\right)\sqrt{x^2-1}}{x^3-3x^2+4+\left(x^2-4\right)\sqrt{x^2-1}}\)
Rút gọn: A=\(\sqrt[3]{\frac{x^3-3x+\left(x^2+1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\).
Với x\(\ge\)2.
Rút gọn:
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\)
rút gọn
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\)
x khác 2,-2
rút gọn
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\)
x khác 2,-2
rút gọn A
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\)
x khác -2,2
Ta có:
A = \(\frac{1x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}\)
= \(\frac{\left(x-2\right)\left(x+1\right)^2+\left(x^2-1\right)\sqrt{x^2-4}}{\left(x+2\right)\left(x-1\right)^2+\left(x^2-1\right)\sqrt{x^2-4}}\)
= \(\frac{\sqrt{x-2}\left(x+1\right)\left(\sqrt{x-2}\left(x+1\right)+\sqrt{x+2}\left(x-1\right)\right)}{\sqrt{x+2}\left(x-1\right)\left(\sqrt{x-2}\left(x+1\right)+\sqrt{x+2}\left(x-1\right)\right)}\)
= \(\frac{\sqrt{x-2}\left(x+1\right)}{\sqrt{x+2}\left(x-1\right)}\)
Với số thực x>=1, x khác 2 hãy rút gọn biểu thức A=\(\frac{^{x^3+3x^2+\left(x^2-4\right)\sqrt{x^2-1-4}}}{x^3+3x^2+\left(x^2-4\right)\sqrt{x^2-1+4}}\)
rút gọn A.
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-4\right)\sqrt{x^2-4}+2}\)
\(\left(x\ne2;-2\right)\)
x + 254 = 245 + 368
x + 254 = 613
x = 613 - 254
x = 359
Rút gọn:
\(\frac{\sqrt[3]{x^3-3x\left(x^2-1\right)\sqrt{x^2-4}}}{2}\) - \(\frac{\sqrt[3]{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}}{2}\)