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![](https://rs.olm.vn/images/avt/0.png?1311)
a)(x-10)2-x(x+80)
(x2-2x10+100)-x2-80x
=x2-20x+100-x2-80x=-100x+100
khi x = 0.98
ta có
(-100*0.98)+100=-98+100=2
b)x3-9x+27x-27
hình như là -27x :))
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-10\right)^2-x\left(x+80\right)\)
\(=x^2-20x+100-x^2-80x\)
\(=-100x+100\)
Thay x=0,98...................................................
b) tương tự phần a
c)\(4x^2-28x+49\)
=\(\left(2x\right)^2-2.2x.7+7^2\)
=(2x-7)2
d) cũng là hằng đăgr thức
a)\(\left(x-10\right)^2-x\cdot\left(x+80\right)\)với x = 0,98
=\(x^2-2\cdot x\cdot10+10^2\)\(-x^2-80x\)
=\(x^2-20x+100-x^2-80x\)
=\(-100x+100\)
=\(-100\cdot0,98+100\)
=\(2\)
b)\(\left(2x+9\right)^2-x\cdot\left(4x+31\right)\)với x=-16,2
=\(\left(2x\right)^2+2\cdot2x\cdot9+9^2-4x^2-31x\)
=\(4x^2+36x+81-4x^2-31x\)
=\(5x+81\)
=\(5\cdot\left(-16,2\right)+81\)
=\(0\)
c)\(4x^2-28x+49\)với x=4
=\(\left(2x\right)^2-2\cdot2x\cdot7+7^2\)
=\(\left(2x-7\right)^2\)
=\(\left(2\cdot4-7\right)^2\)
=\(1\)
Sorry câu d mình không biết
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x-10)^2-x(x+80)
=x2-20x+100-x2-80x
=-100x+100
Thay x=0,98
=-100.0,98+100
=2
b, (2x+9)^2-x(4x+31)
=4x2+36x+81-4x2-31x
=5x+81
Thay x=16,2
=5.16,2+81
=162
c, 4x^2-28x +49
=(2x)2-2.2x7+72
=(2x-7)2
Thay x=4
=(2.4-7)2
=12
=1
đ, x^2-9x^2+27x-27
=k bt lm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(2x-7\right)^2=\left(2.4-7\right)^2=1\)
\(b,\left(x-3\right)^3=\left(5-3\right)^3=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)