Cho 4a2-15ab+3b2=0,b≠4a, b≠-4a. Tính giá trị của biểu thức:T=\(\dfrac{5a-b}{4a-b}\)+\(\dfrac{3b-2a}{4a+b}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(4a^2-15ab+3b^2=0\)
\(4a^2+3b^2=15ab\)
Ta lại có:
\(T=\frac{5a-b}{4a-b}+\frac{3b-2a}{4a+b}\)
\(T=\frac{\left(4a+b\right)\left(5a-b\right)+\left(4a-b\right)\left(3b-a\right)}{16a^2-b^2}\)
\(T=\frac{12a^2+15ab-4b^2}{16a^2-b^2}\)
\(T=\frac{12a^2+4a^2+3b^2-4b^2}{16a^2-b^2}\)
\(T=\frac{16a^2-b^2}{16a^2-b^2}\)
\(T=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Từ\) \(giả\) \(thiết\) : \(4a^2+b^2=\text{5}ab\)
\(\Leftrightarrow4a^2-4ab-ab+b^2\)
\(\Leftrightarrow\left(4a-b\right)\left(a-b\right)=0\)
\(TH1:\) \(4a-b=0\) \((\) \(mẫu\) \(thuẫn\) \(với\) \(2a>b\) \()\)
\(TH2:\) \(a-b=0\)
\(\Rightarrow a=b\)
\(\Rightarrow A=\dfrac{a^2}{4a^2-a^2}\)
\(\Rightarrow A=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)
Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a-b=11\)
\(P=\dfrac{5a-b}{4a+11}+\dfrac{5b-a}{4b-11}=\dfrac{5a-b}{4a+a-b}+\dfrac{5b-a}{4b-\left(a-b\right)}\)
\(=\dfrac{5a-b}{5a-b}+\dfrac{5b-a}{5b-a}\)
\(=2\)
Vậy...