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giải bài toán: cho tam giác MNP, NTlà phân giác của góc N biết MN=4cm, NT=10cm, MP=8cm:TínhTM, TP?
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a: \(3ab-6a^2b\)
\(=3ab\cdot1-3ab\cdot2a\)
=3ab(1-2a)
b: \(x^3-6x\)
\(=x\cdot x^2-x\cdot6\)
\(=x\left(x^2-6\right)\)
c: \(x^2-y^2-9x+9y\)
\(=\left(x^2-y^2\right)-\left(9x-9y\right)\)
\(=\left(x-y\right)\left(x+y\right)-9\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-9\right)\)
d: \(5x^2+10xy+5y^2\)
\(=5\left(x^2+2xy+y^2\right)\)
\(=5\left(x+y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
B = 2 a − 3 a + 1 − a − 4 2 − a a + 7 = 2 a 2 + 2 a – 3 a – 3 – ( a 2 – 8 a + 16 ) – ( a 2 + 7 a ) = 2 a 2 + 2 a – 3 a – 3 – a 2 + 8 a – 16 – a 2 – 7 a = - 19
Đáp án cần chọn là: D
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Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
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a, \(\left(2x-7\right)^2+\left(2x+7\right)^2-2\left(2x+7\right)\left(2x-7\right)=\left(2x-7-2x-7\right)^2=\left(-14\right)^2=196\)
b, \(\left(5x-3\right)^2-\left(5x+3\right)^2-15\left(2x-1\right)\)
\(=\left(5x-3-5x-3\right)\left(5x-3+5x+3\right)-15\left(2x-1\right)\)
\(=-6.10x-15\left(2x-1\right)\)
\(=-60x-15\left(2x-1\right)=-15\left(4x+2x-1\right)=-15\left(6x-1\right)=-90x+15\)
\(=a^3+b^3+3a^2b+3ab^2-a^3+b^3+3a^2b-3ab^2-6a^2b+7-2b^3\)
\(=7\)