Cho B = \(\frac{1-ax}{1+ax}\sqrt{\frac{1+bx}{1-bx}}\) voi a<b<0
va x = \(\frac{1}{a}\sqrt{\frac{2a-b}{b}}\)
RUT GON B
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Cho \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\)với 0<a<b<2a.
Rút gọn P=\(\frac{1+ax}{1-ax}\sqrt{\frac{1-bx}{1+bx}}\)
\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)
Ta có: \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow\text{ax}=\sqrt{\frac{2a-b}{b}}\Rightarrow1+\text{ax}=1+\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}}\)
\(1-\text{ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)
\(\Rightarrow\frac{1+\text{ax}}{1-\text{ax}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}-\sqrt{2a-b}}=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2b-2a}\left(1\right)\)
\(bx=\frac{b}{a}\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}\left(2a-b\right)}{a}\Rightarrow\hept{\begin{cases}1-bx=\frac{a-\sqrt{b\left(2a-b\right)}}{a}\\1+bx=\frac{a+\sqrt{b\left(2a-b\right)}}{a}\end{cases}}\)
\(\Rightarrow\frac{1-bx}{1+bx}=\frac{a-\sqrt{b\left(2a-b\right)}}{a+\sqrt{b\left(2a-b\right)}}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{a^2-2ab+b^2}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{\left(a-b\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)
\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)
Rút gọn biểu thức: \(A=\frac{1-ax}{1+ax}\sqrt{\frac{1+bx}{1-bx}}\) với \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\) và 0<a<b<2a
\(A=\frac{1-ax}{1+ax}.\sqrt{\frac{1+bx}{1-bx}}\) biết a=\(\frac{1}{a}.\sqrt{\frac{2a-b}{b}}\)
với 0<a<b<2a
giúp mình nha các bạn
Tính:
a) \(A=\frac{1-ax}{1+ax}\sqrt{\frac{1+bx}{1-bx}}\) tại \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\)
b) \(B=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
c) \(C=\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}\)
Cho phuong trinh \(ax^2+bx+1=0\) voi a,b la cac so huu ti.Tim a,b biet \(x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)la nghiem cua phuong trinh
1.tính giá trị biểu thức \(A=\frac{1-ax}{1+ã}\sqrt{\frac{1+bx}{1-bx}}\)với \(x=\frac{1}{a}\sqrt{\frac{2a}{b}-1}\left(0< a< b< 2a\right)\)
2. cho \(M=\sqrt{\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1}\).rút gọn M vs \(0\le x\le1\)
biết \(\frac{lim}{x->-\infty}\left(ax+\sqrt{x^2+bx+1}\right)=\frac{1}{2}\) Tính A=2a+b
Tính giá trị của biểu thức: \(A=\dfrac{1-ax}{1+ax}\sqrt{\dfrac{1+bx}{1-bx}}\) với \(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\) (0<a<b<2a)
Tham khảo:
\(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\Rightarrow ax=\sqrt{\dfrac{2a}{b}-1}\)
\(\Rightarrow\left\{{}\begin{matrix}1+ax=\dfrac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}\\1-ax=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1-ax}{1+ax}=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2\left(b-a\right)}\)
Lại có:
\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)
\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)
\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)
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Tính giá trị của biểu thức: \(A=\dfrac{1-ax}{1+ax}\sqrt{\dfrac{1+bx}{1-bx}}\) với \(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\) (0<a<b<2a)