36(x-y)-25(2x-1)2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(36\left(x-y\right)^2-25\left(2x-1\right)^2\)
\(=36\left(y^2-2xy+x^2\right)-25\left(4x^2-4x+1\right)\)
\(=36y^2-72xy+36x^2-100x^2+100x-25\)
\(=36y^2-72xy-64x^2+100x-25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)
\(=\left(x^2+1\right)^2-2\cdot\left(x^2+1\right)\cdot3+3^2\)
\(=\left(x^2+1-3\right)^2\)
\(=\left(x^2-2\right)^2\)
b) Ta có: \(16\left(x+1\right)^2-25\left(2x+3\right)^2\)
\(=\left[4\left(x+1\right)\right]^2-\left[5\left(2x+3\right)\right]^2\)
\(=\left(4x+4\right)^2-\left(10x+15\right)^2\)
\(=\left(4x+4-10x-15\right)\left(4x+4+10x+15\right)\)
\(=\left(-6x-11\right)\left(14x+19\right)\)
c) Ta có: \(x^{16}-1\)
\(=\left(x^8+1\right)\left(x^8-1\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
d) Ta có: \(49\left(x+y\right)^2-36\left(2x+3y\right)^2\)
\(=\left[7\left(x+y\right)\right]^2-\left[6\left(2x+3y\right)\right]^2\)
\(=\left(7x+7y\right)^2-\left(12x+18y\right)^2\)
\(=\left(7x+7y-12x-18y\right)\left(7x+7y+12x+18y\right)\)
\(=\left(-5x-11y\right)\left(19x+25y\right)\)
e) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y\right)^2-2\cdot\left(x+y\right)\cdot1+1^2\)
\(=\left(x+y-1\right)^2\)
f) Ta có: \(x^6-8\)
\(=\left(x^2\right)^3-2^3\)
\(=\left(x^2-2\right)\left(x^4+2x^2+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)
\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)
\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)
Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)
Ủng hộ = 1 cái t i c k nha cảm ơn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(36\left(x-y\right)^2-25\left(2x-1\right)^2=\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left(6x-6y+10x-5\right)\left(6x-6y-10x+5\right)=\left(16x-6y-5\right)\left(-16x-6y+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
36(x-y)2-25(2x-y)2
= 36(x-y)2 - 100(x-y)2
=(36-100)(x-y)2
= -64(x-y)2
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)
2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)
3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)
4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)