K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 11 2021
Câu a bạn sửa lại đề 11→1
\(a,VT=\dfrac{a^2-2a+1}{\left(a-1\right)\left(a^2+1\right)}\cdot\dfrac{a^2+1}{a^2+a+1}\\ =\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{a-1}{a^2+a+1}=VP\)
\(b,=\left[\dfrac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}-x\right]\cdot\dfrac{\left(1+x\right)\left(1-x^2\right)}{1+x}\\ =\dfrac{\left(x^2+1\right)\left(1+x\right)\left(1-x^2\right)}{1+x}=\left(x^2+1\right)\left(1-x^2\right)=VP\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 7 2023
a) A có nghĩa khi: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) \(A=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\left(1+\dfrac{1}{a}\right)\)
\(A=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a}{a}+\dfrac{1}{a}\right)\)
\(A=\left(\dfrac{1-\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\dfrac{1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(A=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)
\(A=\left(\dfrac{-2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right)\cdot\dfrac{a+1}{a}\)
\(A=\left(\dfrac{2}{1-a}-\dfrac{a^2+1}{1-a^2}\right)\cdot\dfrac{a+1}{a}\)
\(A=\left(\dfrac{1+a}{\left(1+a\right)\left(1-a\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right)\cdot\dfrac{a+1}{a}\)
\(A=\left(\dfrac{1+a-a^2-1}{\left(1+a\right)\left(1-a\right)}\right)\cdot\dfrac{a+1}{a}\)
\(A=\dfrac{a-a^2}{\left(1+a\right)\left(1-a\right)}\cdot\dfrac{a+1}{a}\)
\(A=\dfrac{a\left(1-a\right)}{\left(1+a\right)\left(1-a\right)}\cdot\dfrac{a+1}{a}\)
\(A=\dfrac{a}{1+a}\cdot\dfrac{a+1}{a}\)
\(A=\dfrac{a\left(a+1\right)}{a\left(a+1\right)}\)
\(A=1\)
Vậy giá trị của A không phụ thuộc và biến
29 tháng 7 2023
a: ĐKXĐ: a>0; a<>1
b: \(A=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-a\right)}+\dfrac{a^2+1}{a^2-1}\right)\cdot\dfrac{a+1}{a}\)
\(=\left(\dfrac{-2}{2\left(a-1\right)}+\dfrac{a^2+1}{a^2-1}\right)\cdot\dfrac{a+1}{a}\)
\(=\dfrac{-a-1+a^2+1}{\left(a-1\right)\left(a+1\right)}\cdot\dfrac{a+1}{a}\)
\(=\dfrac{a\left(a-1\right)}{a\left(a-1\right)}=1\)
H
1 tháng 4 2021
Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.
a) \(\dfrac{a}{1+b^2}=\dfrac{a\left(1+b^2\right)}{1+b^2}-\dfrac{ab^2}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\)
(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)
\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)
b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)
c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
8 tháng 7 2017
\(a\text{)}.\:\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\\ =\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(x+\sqrt{x}\right)\left(x-\sqrt{x}\right)}+\dfrac{2\sqrt{x}}{x-1}\\ =\dfrac{-2\sqrt{x}}{x\left(x-1\right)}+\dfrac{2\sqrt{x}}{x-1}=\dfrac{-2\sqrt{x}}{x\left(x-1\right)}+\dfrac{2x\sqrt{x}}{x\left(x-1\right)}\\ =\dfrac{2\sqrt{x}\left(x-1\right)}{x\left(x-1\right)}=\dfrac{2\sqrt{x}}{x}=\dfrac{2}{\sqrt{x}}\)
8 tháng 7 2017
\(b\text{)}.\: \left(\dfrac{1}{2\sqrt{a}-a}+\dfrac{1}{2\sqrt{a}+a}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}}\\ =\dfrac{4\sqrt{a}}{4a-a^2}:\dfrac{\sqrt{a}+1}{a-2\sqrt{a}}=\dfrac{4\sqrt{a}}{a\left(4-a\right)}.\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}+1}\\ =\dfrac{4\left(\sqrt{a}-2\right)}{\left(4-a\right)\left(\sqrt{a}+1\right)}=\dfrac{-4\left(2-\sqrt{a}\right)}{\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)\left(\sqrt{a}+1\right)}\\ =-\dfrac{4}{\left(2+\sqrt{a}\right)\left(\sqrt{a}+1\right)}\)
VT
7 tháng 8 2017
bạn tìm trong nâng cao phát triển toán 9 tập 1 ấy nó có ở đấy