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\(M=\left(\frac{2+\sqrt{a}}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\frac{a\left(\sqrt{a}+1\right)-\left(\sqrt{a}+1\right)}{a}\)
\(=\frac{\left(2+\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}-2+a-\sqrt{a}-a-\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a-1}\right)}{a\left(\sqrt{a}+1\right)}=\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)
\(=4\cdot\left(a-1\right)\)
vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Câu C : Lần đầu làm dạng này :))
Xét hiệu A - 2 , ta có :
\(A-2=\frac{2\sqrt{a}+2-4a-2}{2a+1}=\frac{2\sqrt{a}-4a}{2a+1}=\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\)
Ta thấy :
+) Do \(a\ge0\)\(\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)\le0\)
+) a khác 1 ; \(a\ge0\)=> 2a + 1 > 0
\(\Rightarrow\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\le0\)
\(\Leftrightarrow A< 2\)
P/s : sai bỏ qua :))
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-a}\right)\)
ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(A=\left(\frac{\sqrt{a}+1+1-\sqrt{a}}{\sqrt{a}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\frac{2a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(A=\frac{2}{\sqrt{a}-1}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2a+1}\)
\(A=\frac{2\left(\sqrt{a}+1\right)}{2a+1}\)
b) \(a=1-\frac{\sqrt{3}}{2}=\frac{2}{2}-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}\)( tmđk )
Rồi từ đây thế vô :)
c) Nhờ cao nhân làm tiếp chứ em mới lớp 8 thôi ạ :(
a) ĐK: \(0< a< 1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(\sqrt{1+a}-\sqrt{1-a}\right)\left(\sqrt{1+a}+\sqrt{1-a}\right)}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
b) Xét: \(Q^3-Q=\left(a-1\right)^3-\left(a-1\right)=\left(a-1\right)^2\left(a-1-1\right)=\left(a-1\right)^2\left(a-2\right)\)
Do \(a< 1\)=> \(a-2< 0\) và \(a-1< 0\)
nên \(\left(a-1\right)^2\left(a-2\right)< 0\)
=> \(Q^3-Q< 0\)
<=> \(Q^3< Q\)