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Câu b sai kết quả
Kết quả = 1/x nhé
Câu c sai dòng cuối, dòng cuối vầy nè:
= 2(x - 1)/[(x - 1)(x + 1)]
= 2/(x + 1)
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\(d,=\dfrac{3y}{5x\left(x-y\right)}\\ e,=\dfrac{5x\left(x+2\right)\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\dfrac{-5x}{4}\\ f,=\dfrac{3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(6-x\right)}=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\\ g,=\dfrac{3xy\left(x-3y\right)\left(x+3y\right)}{2x^2y^2\left(x-3y\right)}=\dfrac{3\left(x+3y\right)}{2xy}\\ h,=\dfrac{45x^2y\left(x-y\right)\left(x+y\right)}{10xy\left(y-x\right)}=\dfrac{-9x\left(x+y\right)}{2}\\ i,=\dfrac{12\left(a-b\right)\left(a+b\right)\left(a^2+ab+b^2\right)}{3\left(a+b\right)\left(a-b\right)^2}=\dfrac{4\left(a^2+ab+b^2\right)}{a-b}\)
e: \(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=-\dfrac{5}{2}\)
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a) \(A=\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{x+2}\)
Thay \(x=\dfrac{1}{2}\) vào A ta được:
\(A=\dfrac{2\cdot\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{-3}{\dfrac{5}{2}}=-\dfrac{6}{5}\)
b) \(B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
Thay \(x=-5,y=10\) vào B ta đc:
\(B=\dfrac{-5}{-5+10}=-1\)
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d: \(\left(x-y\right)^3-15\left(x-y\right)^2+75\left(x-y\right)-125\)
\(=\left(x-y-5\right)^3\)
c: \(\dfrac{1}{8}x^3+\dfrac{3}{4}x^2y^2+\dfrac{3}{2}xy^4+y^6=\left(\dfrac{1}{2}x+y^2\right)^3\)
chi tiết hơn đc hok bạn mình cũng hok hỉu lắm (mong bạn phản hồi :) :) :) )
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Câu 1. B
Câu 2. C
Câu 3. C
Câu 4. A
Câu 5. B
Câu 6. C
Câu 7. A
Câu 8. D
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chu vi hình vuông là
50x4=200cm
độ dài 1 cạnh hình vuông là
50:4=12,5cm
diện tích hình vuông là
12,5x12,5=156,25cm2
\(a,A=\frac{x^4-5x^2+4}{x^4-10x^2+9}\)
\(A=\frac{x^4-4x^2+4-x^2}{x^4-6x^2+9-4x^2}\)
\(A=\frac{\left(x^2-2\right)^2-x^2}{\left(x^2-3\right)^2-\left(2x\right)^2}\)
\(A=\frac{\left(x^2-2-x\right)\left(x^2-2+x\right)}{\left(x^2-3-2x\right)\left(x^2-3+2x\right)}\)
\(A=\frac{\left(x^2-2x+x-2\right)\left(x^2-x+2x-2\right)}{\left(x^2-3x+x-3\right)\left(x^2-x+3x-3\right)}\)
\(A=\frac{\left(x+1\right)\left(x-2\right)\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x+1\right)\left(x-1\right)\left(x+3\right)}\)
\(A=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x^2-4}{x^2-9}\)
\(b,A=\frac{x^2-4}{x^2-9}=0\)
\(x^2-4=0\)
\(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)vậy có n0 \(x=2;x=-2\)để A=0
câu c mình ko nhìn rõ đề bài
Đk: x \(\ne\)\(\pm\)1; x \(\ne\)\(\pm\)3
A = \(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-4x^2-x^2+4}{x^4-9x^2-x^2+9}=\frac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}==\frac{x^2-4}{x^2-9}\)
b) Với x khác +-1; x khác +-3
Ta có: A = 0 <=> \(\frac{x^2-4}{x^2-9}=0\) =>> \(x^2-4=0\) <=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)