Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
bạn làm rồi nên mk chỉ viết kq thôi nhé :)
a)\(\dfrac{4\sqrt{6x}}{3}\)
b)\(\left(2-y\right)\sqrt{xy}\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
\(a.\sqrt{\left(1-\sqrt{5}\right)^2}+1=\left|1-\sqrt{5}\right|+1=\sqrt{5}-1+1=\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}-2=\sqrt{\left(\sqrt{2}+1\right)^2}-2=\sqrt{2}+1-2=\sqrt{2}-1\)
\(c.\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)=\sqrt{\left(b-\dfrac{1}{2}\right)^2}-2b+\dfrac{1}{2}=b-\dfrac{1}{2}-2b+\dfrac{1}{2}=-2b\)
\(d.\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}\)
\(e.\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-2\)
\(g.3x+\sqrt{x^2-2x+1}=3x+\sqrt{\left(x-1\right)^2}\)
* \(x\ge1\Rightarrow3x+\left|x-1\right|=3x+x-1=4x-1\)
* \(x< 1\Rightarrow3x+\left|x-1\right|=3x+1-x=2x+1\)
\(h.\sqrt{y+2\sqrt{y^2-2y+1}}=\sqrt{y+2\sqrt{\left(y-1\right)^2}}=\sqrt{y+2y-2}=\sqrt{3y-2}\left(y\ge1\right)\) hoặc: \(\sqrt{y+2-2y}=\sqrt{-y+2}\left(y< 1\right)\)
\(H=\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
\(H^2=17-2\sqrt{32}+17+2\sqrt{32}+2\sqrt{\left(17-2\sqrt{32}\right)\left(17+2\sqrt{32}\right)}=34+2\sqrt{161}\)
\(H=\sqrt{34+2\sqrt{161}}\)
\(k.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
bài 2: ta có : \(Q=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-\left(1-a\right)}\right)\left(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\right).\sqrt{a^2-2a+1}\)
\(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}\sqrt{1-a}+1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\dfrac{\sqrt{1-a^2}}{a}-\dfrac{1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1-a^2}+1}{a}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{1-a^2-1}{a^2}\right)\left(1-a\right)=a-1\)b) ta có : \(Q^3-Q=\left(a-1\right)\left(\left(a-1\right)^2-1\right)=a\left(a-1\right)\left(a-2\right)\)
mà ta có : \(\left\{{}\begin{matrix}a>0\\a-1< 0\\a-2< 0\end{matrix}\right.\Rightarrow a\left(a-1\right)\left(a-2\right)>0\) \(\Rightarrow Q^3-Q>0\Leftrightarrow Q^3>Q\)
vậy \(Q^3>Q\)
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh Hằngsoyeon_Tiểubàng giảiMashiro ShiinaVõ Đông Anh Tuấn
Hoàng Lê Bảo NgọcTrần Việt Linh
cứu tôi với
Bài 1)
ĐK: \(x\geq 0; x\neq -4\)
Ta có:
\(A=\frac{1}{\sqrt{x}+2}+\frac{1}{2+\sqrt{x}}-\frac{2\sqrt{x}}{x+4}\)
\(=\frac{2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x+4}=2\left(\frac{1}{\sqrt{x}+2}-\frac{\sqrt{x}}{x+4}\right)\)
\(=2.\frac{x+4-x-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=2.\frac{4-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=\frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}\)
\(B=(\sqrt{2}+\sqrt{3}).\sqrt{2}-\sqrt{6}+\frac{\sqrt{333}}{\sqrt{111}}\)
\(=2+\sqrt{6}-\sqrt{6}+\frac{\sqrt{3}.\sqrt{111}}{\sqrt{111}}=2+\sqrt{3}\)
Để \(A=B\Leftrightarrow \frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}=2+\sqrt{3}\)
PT rất xấu. Mình nghĩ bạn đã chép sai biểu thức A.
Bài 2 : Tọa độ điểm B ?
Bài 3:
Để pt có hai nghiệm thì \(\Delta'=(m-3)^2-(m^2-1)>0\)
\(\Leftrightarrow 10-6m>0\Leftrightarrow m< \frac{5}{3}\)
Áp dụng định lý Viete: \(\left\{\begin{matrix} x_1+x_2=2(m-3)\\ x_1x_2=m^2-1\end{matrix}\right.\)
Khi đó:
\(4=2x_1+x_2=x_1+(x_1+x_2)=x_1+2(m-3)\)
\(\Rightarrow x_1=10-2m\)
\(\Rightarrow x_2=2(m-3)-(10-2m)=4m-16\)
Suy ra: \(\Rightarrow x_1x_2=(10-2m)(4m-16)\)
\(\Leftrightarrow m^2-1=8(5-m)(m-4)\)
\(\Leftrightarrow m^2-1=8(-m^2+9m-20)\)
\(\Leftrightarrow 9m^2-72m+159=0\)
\(\Leftrightarrow (3m-12)^2+15=0\) (vô lý)
Vậy không tồn tại $m$ thỏa mãn điều kiện trên.
a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)
b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)
\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)
c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)
\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)