bài 1, rút gọn và tìm giá trị của x tại a=5,b=2
A=\(\frac{a^2+\text{ã}+ab+bx}{a^2+\text{ax}-ab-bx}\)
K
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29 tháng 11 2016
1, b) \(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\) = \(\frac{\left(x^2+2xy+y^2\right)-4}{\left(x^2+4x+4\right)-y^2}\) =\(\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)= \(\frac{\left(x+y+2\right)\left(x+y-2\right)}{\left(x+2+y\right)\left(x+2-y\right)}\) = \(\frac{x+y-2}{x+2-y}\)
2, A= \(\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\) = \(\frac{\left(a^2+ax\right)+\left(ab+bx\right)}{\left(a^2+ax\right)-\left(ab+bx\right)}\) = \(\frac{a\left(a+x\right)+b\left(a+x\right)}{a\left(a+x\right)-b\left(a+x\right)}\)= \(\frac{\left(a+x\right)\left(a+b\right)}{\left(a+x\right)\left(a-b\right)}\)= \(\frac{a+b}{a-b}\)
NL
1
21 tháng 12 2016
\(A=\frac{x^2+ax+bx+ab}{x}\) đúng thế không?
DK x khác 0
\(A=x+a+b+\frac{ab}{x}=\left(\sqrt{x}-\sqrt{\frac{ab}{x}}\right)^{^2}+\left(\sqrt{a}+\sqrt{b}\right)^2\)
Amin =\(\left(\sqrt{a}+\sqrt{b}\right)^2\)
KHi \(\sqrt{x}=\sqrt{\frac{ab}{x}}\Leftrightarrow x=\sqrt{ab}\)
17 tháng 8 2020
\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)
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) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)
\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)
12 tháng 9 2021
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:
\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)
\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)
\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)
\(A=\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\)
\(=\frac{a\left(a+b\right)+x\left(a+b\right)}{a\left(a-b\right)+x\left(a-b\right)}\)
\(=\frac{\left(a+b\right)\left(a+x\right)}{\left(a-b\right)\left(a+x\right)}\)
\(=\frac{a+b}{a-b}\)
Thay \(a=5;b=2\) vào A ta có:
\(A=\frac{5+2}{5-2}=\frac{7}{3}\)
Vậy tại \(a=5;b=2\) thì A=7/3