HOC24
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\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(\left(b+c\right)\left(c-a\right)+\left(c+a\right)\left(b-c\right)\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(2bc-2ac\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)} =\dfrac{-2c\left(a+b\right)\left(a-b\right)+\left(a-b\right)\left(b+c\right)\left(c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(-2ac-2bc+bc+ab+c^2+ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(ab-bc-ac+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
\(C=\dfrac{9x^2-16}{3x^2-4x}\) ĐKXĐ: 3x2-4x\(\ne\)0 \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{4}{3}\end{matrix}\right.\) C=0 <=> x=4/3(ktm đkxđ) hoặc x=-4/3(tm đkxđ) =>x=-4/3 \(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}=3+\dfrac{4}{x}\) Thôi câu khác tương tự nhá
\(B=\dfrac{x^2-9}{x^2-6x+9}=\dfrac{x^2-9}{\left(x-3\right)^2}\) ĐKXĐ: (x-3)2 \(\ne\)0 <=> \(x\ne3\) B=0 <=> x2-9=0 <=> x=3(ko tm đkxđ) or x=-3(tm đkxđ)=>x=-3 \(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
\(A=\dfrac{2x+6}{\left(x+3\right)\left(x-2\right)}\) ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\) A=0 <=> 2x+6=0 <=> x=-3(ko tm đkxđ) => ko có x để A=0 \(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
Đề sai sai kiểu j ấy, ko có quy luật j cả
Nhìu thế này khó mà làm hết