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`(x-3)^2 - x^2 + 10x - 7`
`= x^2 - 6x + 9 - x^2 + 10x - 7`
`= (x^2 - x^2) + (-6x + 10x) + (9-7)`
`= 4x + 2`
\(\left(x-3\right)^2-x^2+10x-7\)
\(=\left(x^2-6x+9\right)-x^2+10x-7\)
\(=\left(x^2-x^2\right)-\left(6x-10x\right)+\left(9-7\right)\)
\(=4x+2\)
\(=2\left(2x+1\right)\)
a) (x2 – x) . (2x2 – x – 10)
= x2 . (2x2 – x – 10) – x. (2x2 – x – 10)
= x2 . 2x2 + x2 . (-x) + x2 .(-10) – [ x. 2x2 + x. (-x) + x. (-10)]
= 2x4 – x3 - 10x2 – (2x3 – x2 – 10x)
= 2x4 – x3 - 10x2 – 2x3 + x2 + 10x
= 2x4 + (– x3 – 2x3 ) + (-10x2 + x2 )+ 10x
= 2x4 – 3x3 - 9x2 + 10x
b) (0,2x2 – 3x) . 5(x2 -7x + 3)
= (0,2x2 . 5 – 3x . 5) . (x2 -7x + 3)
= (x2 – 15x). (x2 -7x + 3)
= x2 . (x2 -7x + 3) – 15x. (x2 -7x + 3)
= x2 . x2 + x2 . (-7x) + x2 . 3 – [ 15x3 + 15x.(-7x) + 15x.3]
= x4 – 7x3 + 3x2 – (15x3 – 105x2 + 45x)
= x4 – 7x3 + 3x2 – 15x3 + 105x2 – 45x
= x4 +(– 7x3 – 15x3 )+ (3x2 + 105x2) – 45x
= x4 – 22x3 + 108x2 – 45x
a: A=x^5-32
Khi x=3 thì A=3^5-32=243-32=211
b: B=x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+x^7-x^6+x^5-x^4+x^3-x^2+x-1
=x^8-1
=2^8-1=255
a)
b) (x – 1)(x + 1)(x2 + 1)
= [x .(x + 1) – 1 .(x + 1)] . (x2 + 1)
= {x.x + x.1 + (-1).x + (-1).1}. (x2 + 1)
= (x2 + x – x – 1) . (x2 + 1)
= (x2 – 1) . (x2 + 1)
= x2 . (x2 +1) – 1.(x2 + 1)
= x2 . x2 + x2 . 1 – (1.x2 + 1.1)
= x4 + x2 – (x2 + 1)
= x4 + x2 – x2 – 1
= x4 – 1
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x(x2 – y) – x2 (x + y) + y (x2– x) = x3 – xy – x3 – x2y + yx2 – yx= (2x-2y) – (x2 -2xy +y2) =2(x-y) – (x-y)2
Với x =1/2, y = -100 biểu thức có giá trị là -2 . 1/2. (-100) = 100.
a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)
\(=5x^3-5x^2y+5x-2x^2y+2xy^2-2y\)
\(=5x^3-7x^2y+2xy^2+5x-2y\)
b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
(\(x\) - 3).(3 + \(x\))
= 3\(x\) + \(x^2\) - 9 - 3\(x\)
= \(x^2\) - 9
\(\left(x-3\right)\left(3+x\right)\\ =\left(x-3\right).3+\left(x-3\right).x\\ =3x-9+2x-3x\\ =3x-9-x\\ =3x-x+9\\ =2x+9.\)
\(\left(x-\dfrac{1}{2}\right)^3=81=\left(\sqrt[3]{81}\right)^3\)
\(\Leftrightarrow x-\dfrac{1}{2}=\sqrt[3]{81}\)
\(\Leftrightarrow x=\sqrt[3]{81}+\dfrac{1}{2}\)
`#3107.101107`
`(x - 1/2)^2 = 81?`
`=> (x - 1/2)^2 = (+-9)^2`
`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{2}=9\\x-\dfrac{1}{2}=-9\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=9+\dfrac{1}{2}\\x=-9+\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{17}{2}\end{matrix}\right.\)
Vậy, `x \in {-17/2; 19/2}.`
\(\left(x^3\right)^2:\left(x^2\right)^3\) = \(x^6:x^6=1\)