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Bài 1:
a^2-5ab-6b^2=0
=>a^2-6ab+ab-6b^2=0
=>a*(a-6b)+b(a-6b)=0
=>(a-6b)(a+b)=0
=>a=-b hoặc a=6b
TH1: a=-b
\(A=\dfrac{-2b-b}{-3b-b}+\dfrac{5b+b}{-3b+b}=\dfrac{-3}{-4}+\dfrac{6}{-2}=\dfrac{3}{4}-3=-\dfrac{9}{4}\)
TH2: a=6b
\(A=\dfrac{12b-b}{18b-b}+\dfrac{5b-6b}{18b+b}=\dfrac{11}{17}+\dfrac{-1}{19}=\dfrac{192}{323}\)
Ta có:
\(4a^2+b^2=5ab\Leftrightarrow4a^2+b^2-4ab-ab=0\)
\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\4a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\left(ktm\right)\\4a=b\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow4a=b\)
\(\Rightarrow\dfrac{5ab}{3a^2+2b^2}=\dfrac{5a.4a}{3a^2+2.\left(4a\right)^2}=\dfrac{20a^2}{3a^2+32a^2}\)
\(=\dfrac{20a^2}{35a^2}=\dfrac{4}{7}\)
\(4a^2+b^2=5ab\)
\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)
\(\Rightarrow\left(a-b\right)\left(4a-b\right)=0\)
\(\Rightarrow b=4a\left(do.a\ne b\right)\)
\(\dfrac{5ab}{3a^2+2b^2}=\dfrac{20a^2}{3a^2+32a^2}=\dfrac{4}{7}\)
`a^2+4ab-5b^2=0`
`<=>a^2+4ab+4b^2-9b^2=0`
`<=>(a+2b)^2-9b^2=0`
`<=>(a+2b-3b)(a+2b+3b)=0`
`<=>(a-b)(a+5b)=0`
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-5b\end{matrix}\right.\)
`Q={2a-b}/{a-b}+{3a-2b}/{a+b}`
Với `a=b` `=>` giá trị vô nghĩa
Với `a=-5b`
`Q={-10b-b}/{-5b-b}+{-15b-2b}/{-5b+b}`
`Q={-11b}/{-6b}+{-17b}/{-4b}`
`Q=11/6+17/4`
`Q=73/12`
Ta có \(6a^2-15ab+5b^2=0\Leftrightarrow15ab=6a^2+5b^2\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{9a^2-b^2}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}=\dfrac{3a^2+6a^2+5b^2-6b^2}{9a^2-b^2}\)
\(Q=\dfrac{9a^2-b^2}{9a^2-b^2}=1\)
Ta có:
\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}+\dfrac{\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}\)
Ta lại có:
\(6a^2-15ab+5b^2=0\)
\(\Rightarrow3a^2+15ab-6b^2=9a^2-b^2\left(1\right)\)
Thay (1) vào Q
=> Q = 1
\(\left\{{}\begin{matrix}a.b\ne0\left(!\right)\\9a^2-b\ne0\left(!!\right)\\10a^2-3b^2-5ab=0\left(1\right)\\A=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}-3\left(2\right)\end{matrix}\right.\)
Từ (!) \(\Rightarrow\left(1\right)\Leftrightarrow10-3\left(\dfrac{b}{a}\right)^2-5\left(\dfrac{b}{a}\right)=0\)(3)
Đặt b/a =x
\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x-10=0\\\left[{}\begin{matrix}x_1=\dfrac{-5-\sqrt{5.29}}{6}\\x_2=\dfrac{-5+\sqrt{5.29}}{6}\end{matrix}\right.\end{matrix}\right.\)(4)
Từ (!) \(\Rightarrow\left(2\right)\Leftrightarrow A=\dfrac{2-x}{3-x}+\dfrac{5x-1}{3+x}-3=\left(1-\dfrac{1}{3-x}\right)+\left(5-\dfrac{16}{x+3}\right)-3=B+3\)
\(B=\dfrac{1}{x-3}-\dfrac{16}{x+3}=\dfrac{x+3-16x+48}{x^2-9}=\dfrac{-15x+51}{x^2-9}=\dfrac{3\left(17-5x\right)}{x^2-9}\)
Từ (4)\(\Rightarrow\left\{{}\begin{matrix}17-5x=3x^2+7\\B=\dfrac{3\left(3x^2+7\right)}{x^2-9}\end{matrix}\right.\) \(B=9+\dfrac{81+27}{x^2-9}\)
\(A=12+\dfrac{108}{x^2-9}\)
Bạn tự thay vào :\(\begin{matrix}A\left(x_1\right)=\\A\left(x_2\right)=\end{matrix}\) chú ý bp => x^2 --> mới thay vào
Mình nghi đề của bạn nhầm dấu: biểu thức (1)
\(10a^2-3b^2-5ab=0\Rightarrow10\left(a-\dfrac{b}{4}\right)^2-\dfrac{29b^2}{8}=0\)
\(\Rightarrow a=b=0\)
tự làm tiếp nhé, phần khó nhất mk đã giúp bn r`h thay vào thôi