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Đề sai sửa luôn !
\(a,M=\left(\frac{21}{x^2-9}+\frac{4-x}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21-\left(4-x\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+3-1}{x+3}\right)\)
\(=\frac{21-4x-12+x^2+3x-x^2+3x+x-3}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
\(b,x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Kết hợp ĐKXĐ => x = 2
Thay vào \(M=\frac{3}{2-3}=\frac{3}{-1}=-3\)
Vậy ...........................
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
Bài 1:
a) Rút gọn:
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right).\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{\left(3-x\right).\left(x+3\right)^2}{\left(x+3\right).\left(x-3\right).\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right).\left(x+3\right)^2}{\left(x+3\right)^2.\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-1.\left(x+3\right)}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3+x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}:\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}.\frac{x+3}{3x^2}\)
\(A=\frac{-3.\left(x+3\right)}{\left(x+3\right).3x^2}\)
\(A=\frac{-1}{x^2}.\)
b) Ta có:
\(\left|x\right|=-\frac{1}{2}\)
Vì \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x\right|>-\frac{1}{2}\)
\(\Rightarrow\left|x\right|\ne-\frac{1}{2}\)
\(\Rightarrow x\in\varnothing.\)
Vậy biểu thức A không có giá trị tại \(\left|x\right|=-\frac{1}{2}.\)
Chúc bạn học tốt!
Bài 1: Cho biểu thức: \(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
a) Rút gọn biểu thức \(A\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right)\left(x+3\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3}{x+3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}:\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(A=\frac{-3\left(x+3\right)}{\left(x+3\right)3x^2}\)
\(A=\frac{-1}{x^2}\)