rút gọn
A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)
rút gọn
A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)
\(A=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}=\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}=\dfrac{2a^2+4}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)
\(=\dfrac{2a^2+4}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2a^2+4-\left(1+a+a^2\right)\left[\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)\right]}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2a^2+4-2\left(1+a+a^2\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}\)
\(=\dfrac{2a^2+4-2-2a-2a^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2-2a}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{2}{1+a+a^2}\)
A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)
=\(\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)=
\(\dfrac{2a^2+4}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)
= \(\dfrac{2a^2+4-\left(1-\sqrt{a}+a-a\sqrt{a}+a^2-a^2\sqrt{a}\right)-\left(1+\sqrt{a}+a+a\sqrt{a}+a^2+a^2\sqrt{2}\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)
=\(\dfrac{-2a-2}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{1+a+a^2}\)
rút gọn P=\(\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x-2}}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
Sửa đề: \(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{3\sqrt{x}+3+\sqrt{x}-3}{x-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
tìm max \(\dfrac{1}{\sqrt{x}-1}\)
\(\dfrac{1}{\sqrt{x}-1}\) đạt GTLN khi và chỉ khi \(\sqrt{x}-1\) đạt GTNN
\(\sqrt{x}-1\) đạt GTNN khi \(\sqrt{x}\) đạt GTNN
\(\sqrt{x}\) đạt GTNN khi x=0 (vì \(\sqrt{x}\ge0\) )đây là bài tìm min chứ k phải max, lập luận như cj mai là đúng, min = -1,
cj thảo xem lại đề bài
Tìm x
a) \(\left(2-\sqrt{x}\right)\)\(.\)\(\left(1+\sqrt{x}\right)=\) \(-x+\sqrt{5}\)
b) \(\sqrt{x^2+2x\sqrt{3}+3}\)\(=\) \(\sqrt{3}+x\)
c) \(\sqrt{x^2-2x+4}\)\(=\) \(x+2\)
GIÚP MÌNH VS MAI MÌNH NỘP BÀI RỒI
c) \(\sqrt{\left(x-2\right)^2}=x+2\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x< 2\\-\left(x-2\right)=x+2\Rightarrow2x=0\Rightarrow x=0\end{matrix}\right.\\\left\{{}\begin{matrix}2\le x\\\left(x-2\right)=x+2\Rightarrow-2=2\Rightarrow voN_0\end{matrix}\right.\end{matrix}\right.\)
x=0 duy nhat
Tính
a) \(\dfrac{\sqrt{3}+2}{\sqrt{3}-2}\)\(-\)\(\dfrac{\sqrt{3}-2}{\sqrt{3}+2}\)\(+\)\(\dfrac{8\sqrt{6}-8\sqrt{3}}{\sqrt{2}-1}\)
b) \(\sqrt{\sqrt{2}+2\sqrt{2}-1}\)\(+\)\(\sqrt{\sqrt{2}-2\sqrt{2}-1}\)
b)
\(\sqrt{2}< 2\sqrt{2}=>\sqrt{2}-2\sqrt{2}< 0\)
Rút Gọn
a) A = \(\dfrac{1}{\sqrt{x}+\sqrt{x-1}}\) \(-\)\(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}\)\(-\)\(\dfrac{x\sqrt{x}-x}{1-\sqrt{x}}\)( Rút gọn và Tìm x để A > 0 )
b) B \(=\)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\)\(.\)\(\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\)
c) C = \(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+2\right)\)\(.\)\(\left(2-\dfrac{\sqrt{x}+x}{1+\sqrt{x}}\right)\)
GIÚP MÌNH VS MAI MÌNH NỘP BÀI RỒI
b: \(B=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\cdot\left(2-\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{-\left(5-\sqrt{b}\right)}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
c: \(C=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+2\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\)
=4-x
Giải các phương trình sau
a) \(-x^2+4\cdot x+1=2\cdot\sqrt{2\cdot x+1}\)
b) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
c) \(5\cdot x^2-2\cdot x+1=\left(4\cdot x-1\right)\cdot\sqrt{x^2+1}\)
d) \(\left(2\cdot x-1\right)\cdot\sqrt{10-4\cdot x^2}=5-2\cdot x\)
e) \(\sqrt{2\cdot x-1}-\sqrt{x+1}=2\cdot x-4\)
f) \(\sqrt{x^2-2\cdot x}+\sqrt{2\cdot x^2+4\cdot x}=2\cdot x\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
m bằng mấy để phương trình \(2x-2m\sqrt{x}+m^2-2=0\) có hai nghiệm phân biệt.
Lời giải:
Đặt \(\sqrt{x}=t(t>0)\Rightarrow 2t^2-2mt+m^2-2=0\)
Để pt có hai nghiệm phân biệt thì trước tiên thì:
\(\Delta'=m^2-2(m^2-2)>0\Leftrightarrow 4-m^2>0\)
\(\Leftrightarrow -2< m<2\)
Mặt khác lưu ý rằng hai nghiệm của pt phải đều là nghiệm không âm.
Để đạt được điều ấy thì:
\(\left\{\begin{matrix} t_1+t_2=m>0\\ t_1t_2=\frac{m^2-2}{2}>0\end{matrix}\right.\) (hệ thức Viete)
\(\Leftrightarrow m>\sqrt{2}\)
Vậy \(2> m> \sqrt{2}\)
phân tích ra hàng đẳng thức hộ mk vs
\(\sqrt{54-18\sqrt{2}}\)
\(\sqrt{54-18\sqrt{2}}\) = \(\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.3+3^2}\)=\(\sqrt{\left(3\sqrt{2}+3\right)^2}\)= 3\(\sqrt{2}\)+3
Giúp mình với mình sắp đi học rồi 😭😭😭
câu 21
làm một con có vẻ rắc rối nhất ví dụ thôi
\(\dfrac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\dfrac{\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{5}+\sqrt{2}}=1\)\(=\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{5}+\sqrt{2}}=1\)