H= (1+ \(\dfrac{a
+\sqrt{a}}{\sqrt{â+1}}\)).(1+\(\dfrac{â-\sqrt{a}}{1-\sqrt{â}}\))
B=(1+\(\dfrac{1a+\sqrt{â}}{\sqrt{\text{a}+1}a}\))+(1+\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\))
P=\(\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{â}+1}\)
Rút gọn và tìm x khi P=\(\dfrac{1}{2}\)
Mng giúp mik vs ạ!
a: Ta có: \(P=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{a+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{\left(a+1\right)\cdot\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
Cho biểu thức P= (\(\dfrac{2}{\sqrt{1+a}}\)+ \(\sqrt{1-â}\)) : (\(\dfrac{2}{\sqrt{1-a^2}}\) +1)
a, rút gọn p
b, tính p khi a = 24/49
c, tính a để p=2
\(a,P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\left(-1< a< 1\right)\\ P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{2+\sqrt{\left(1-a\right)\left(1+a\right)}}\\ P=\sqrt{1-a}\\ b,a=\dfrac{24}{49}\Leftrightarrow1-a=\dfrac{25}{49}\\ \Leftrightarrow P=\sqrt{1-a}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\\ c,P=2\Leftrightarrow1-a=4\Leftrightarrow a=-3\left(ktm\right)\Leftrightarrow a\in\varnothing\)
Câu 80**: Tam giác ABC có Â = 1200 , AB = AC, BC = 12 . Độ dài đường cao AH là:
A. \(\sqrt{3}\); B . \(\dfrac{\sqrt{3}+1}{2}\) ; C . \(\dfrac{2+\sqrt{3}}{2}\); D.\(2\sqrt{3}\) .
AB=AC \(\Rightarrow\Delta ABC\) cân tại A
\(\Rightarrow AH\) đồng thời là phân giác và trung tuyến
\(\Rightarrow\left\{{}\begin{matrix}\widehat{BAH}=\dfrac{1}{2}\widehat{A}=60^0\\BH=\dfrac{1}{2}BC=6\end{matrix}\right.\)
Trong tam giác vuông ABH:
\(tan\widehat{BAH}=\dfrac{BH}{AH}\Rightarrow AH=\dfrac{BH}{tan\widehat{BAH}}=\dfrac{6}{tan60^0}=2\sqrt{3}\)
Câu 80**: Tam giác ABC có Â = 1200 , AB = AC, BC = 12 . Độ dài đường cao AH là:
A. √3; B \(\dfrac{\sqrt{3}+1}{2}\). ; C \(\dfrac{2+\sqrt{3}}{2}\).; D\(2\sqrt{3}\). .
giải hộ mik với
Rút gọn các biểu thức sau:
a) \(\frac{x^2-3}{x+\sqrt{3}}\)
b) \(\frac{1-â\sqrt{â}}{1-\sqrt{â}}\) với a\(\ge\) 0 và a\(\ne\) 1
a) \(=\frac{x^2-\sqrt{3^2}}{x+\sqrt{3}}=\frac{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}=a+\sqrt{a+1}\)
Rút gọn biểu thức:
A= \(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{â+\sqrt{a}}+\left[\sqrt{a}-\frac{1}{\sqrt{a}}\right]\left[\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right]\)
P= \(\left(1+\frac{a\sqrt{a}}{\sqrt{a}+1}\right)\times\left(1+\frac{a-\sqrt{a}}{-1+\sqrt{a}}\right)\)
â/ Rút gọn
b/ Tìm a biết P > \(-\sqrt{2}\)
c/ Tìm a biết P = \(\sqrt{a}\)
Giúp m vs.
a) \(^{â^3-2\sqrt{2}}\)
b)\(\sqrt{21}+\sqrt{3}+\sqrt{7}+1\)
c) \(x+2\sqrt{x}-3\)
a) \(a^3-2\sqrt{2}\)
\(=a^3-\left(\sqrt{2}\right)^3\)
\(=\left(a-\sqrt{2}\right)\left(a+a\sqrt{2}+2\right)\)
b) \(\sqrt{21}+\sqrt{3}+\sqrt{7}+1\)
\(=\sqrt{3}\left(\sqrt{7}+1\right)+\left(\sqrt{7}+1\right)\)
\(=\left(\sqrt{7}+1\right)\left(\sqrt{3}+1\right)\)
c) \(x+2\sqrt{x}-3\)
\(=x-\sqrt{x}+3\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)\)