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Tử = \(a^3-3a+2=a^3-1-3a+3\)
\(=\left(a-1\right)\left(a^2+a+1\right)-3\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2+a-2\right)\)
\(=\left(a-1\right)\left(a-1\right)\left(a+2\right)=\left(a-1\right)^2\left(a+2\right)\)
Mẫu =\(2a^3-7a^2+8a-3=2a\left(a^2-2a+1\right)-3\left(a^2-2a+1\right)\)
\(=\left(a-1\right)^2\left(2a-3\right)\)
=>\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}=\frac{\left(a-1\right)^2\left(a+2\right)}{\left(a-1\right)^2\left(2a-3\right)}=\frac{a+2}{2a-3}\)
Nhớ h cho mik nhé
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
Đây là câu a/
https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228
Còn câu b thì như sau:
Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
\(\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a^2-3a}\right):\dfrac{2a+2}{a}=\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9}{a\left(a-3\right)}-\dfrac{a^2}{a\left(a-3\right)}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9-a^2+9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\dfrac{18-6a}{a\left(a-3\right)}:\dfrac{2a+2}{a}=\dfrac{6a-18}{\left(-a\right)\left(3-a\right)}:\dfrac{2a+2}{a}=\dfrac{6}{\left(-a\right)}:\dfrac{2a+2}{a}=\dfrac{6a}{\left(-2a^2\right)+\left(-2a\right)}.DKXD:a\ne0;a\ne3\)
Giải tắt quá bạn ơi!