em làm luôn
\(P=\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{x-1}\)
b) thì em chưa làm đc :((
b, \(x=24-16\sqrt{2}=24-2.8.\sqrt{2}=24-8\sqrt{8}\)
\(=24-2.4\sqrt{8}=4^2-2.4\sqrt{8}+\left(\sqrt{8}\right)^2=\left(4-\sqrt{8}\right)^2\)
*, làm tiếp bước Q làm : \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4-\sqrt{8}\right|=4-\sqrt{8}\)( vì \(4-\sqrt{8}>0\))
hay \(\frac{1}{4-\sqrt{8}-1}=\frac{1}{3-\sqrt{8}}=3+\sqrt{8}\)
Vậy với \(x=24-16\sqrt{2}\)thì \(P=3+\sqrt{8}\)
a) P=1√x−1 .
b) Chú ý √x=√(4−2√2)2=4−2√2.
Vậy P=14−2√2−1 =13−2√2 =32−(2√2)23−2√2 =3+2√2.
Câu hỏi thuộc chủ đề: Rút gọn biểu thức chứa căn bậc hai
o l m . v n
a, \(\dfrac{1}{\sqrt{x}+1}\)
b,
P=\(\dfrac{1}{\sqrt{x}-1}\)
P=3+2\(\sqrt{2}\)
a, P=\(\dfrac{1}{\sqrt{x}-1}\) với x\(\ge0\) ; x\(\ne1\) b, P=\(\dfrac{1}{3-2\sqrt{2}}\) khi x=\(24-16\sqrt{2}\)
a/ ĐKXĐ: x≥0,x≠1
P=\(\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)=\(\dfrac{1}{\sqrt{x}-1}\)
Vậy P=\(\dfrac{1}{\sqrt{x}-1}\) với x≥0,x≠1
b/ ta có x =24−\(16\sqrt{2}\)=24-\(8\sqrt{8}\)=\(4^2-8\sqrt{8}-\sqrt{8^2}\)=\(\left(4-\sqrt{8}\right)^2\)
Thay x =\(\left(4-\sqrt{8}\right)^2\) vào biểu thức P:
P= \(\dfrac{1}{\sqrt{\text{}\text{}\text{}\left(4-\sqrt{8}\right)^2}-1}=\dfrac{1}{3-2\sqrt{2}}\)
Vậy P=\(\dfrac{1}{3-2\sqrt{2}}\) khi x =24−\(16\sqrt{2}\)
a) P= \(\dfrac{1}{\sqrt{x}-1}\)
b) P=3+2\(\sqrt{2}\)
a, P = \(\dfrac{1}{\sqrt{x}-1}\)
b, Thay x = \(24-16\sqrt{2}\) vào bt ,vậy
P = \(\dfrac{1}{3-2\sqrt{2}}\)
P = \(3+2\sqrt{2}\)
Với x≥0;x≠1,ta có:
P=\(\dfrac{3(\sqrt{x}-1)}{(\sqrt{x}+1)\times(\sqrt{x}-1)}-\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)\times(\sqrt{x}-1)}-\dfrac{\sqrt{x}-5}{(\sqrt{x}-1)\times(\sqrt{x}+1)}\)
=\(\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{(\sqrt{x}+1)\times(\sqrt{x}-1)}\)
=\(\dfrac{(\sqrt{x}+1)}{(\sqrt{x}+1)\times(\sqrt{x}-1)}\)
P=\(\dfrac{1}{\sqrt{x}-1}\)
Vậy với x≥0;x≠1 thì P=\(\dfrac{1}{\sqrt{x}-1}\)
b) x=\(24-16\sqrt{2}\)
⇔x=\((16-2\times4\times2\sqrt{2}+8)\)
⇔x=\((4-2\sqrt{2})^2\)
⇔\(\sqrt{x}=\sqrt{(4-2\sqrt{2})^2}\)
⇔\(\sqrt{x}=4-2\sqrt{2}\) (vì 4>2\(\sqrt{2})\)
Thay \(\sqrt{x}=4-2\sqrt{2}\) vào P , ta có
P=\(\dfrac{1}{4-2\sqrt{2}-1}=\dfrac{1}{3-2\sqrt{2}}=3+2\sqrt{2}\)
Vậy P =3+\(2\sqrt{2}\) khi x=\(24+16\sqrt{2}\)
\(P=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}-1}\)
\(a,P=\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
P=\(\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
P=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
P=\(\dfrac{1}{\sqrt{x}-1}\)
b,\(x=\left(4-2\sqrt{2}\right)\left(4-2\sqrt{2}\right)\)
\(\sqrt{x}=4-2\sqrt{2}\)
Thay \(\sqrt{x}=4-2\sqrt{2}\)vào P
P=\(\dfrac{1}{3-2\sqrt{2}}\)
P=\(\dfrac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}\)
a) P=\(\dfrac{1}{\sqrt{x}-1}\)
b) \(\sqrt{x}=\sqrt{\left(4-2\sqrt{2}\right)^2}=4-2\sqrt{2}\)
Vậy P=\(\dfrac{1}{4-2\sqrt{2}-1}=\dfrac{1}{3-2\sqrt{2}}=\dfrac{3^2-\left(2\sqrt{2}\right)^2}{3-2\sqrt{2}}=3+2\sqrt{2}\)
a) P=\(\dfrac{1}{\sqrt{x}-1}\)
b) P=\(3+2\sqrt{2}\)
