Cho đa thức
A=4x2-5x+3;B=3x2+2x-1
Tính A+B;A-B
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a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`
`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`
`=2x^2-x-2x+5/2+4x^2`
`=6x^2-3x+5/2`
b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`
`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`
`=3x-y-y-x+2x^2+2x`
`=2x^2+4x-2y`
\(a,A=y^2-\dfrac{1}{2}y+\dfrac{1}{16}\)
\(=y^2-2.y.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)
\(=\left(y-\dfrac{1}{4}\right)^2\)
Với \(y=100,25\), ta được:
\(A=\left(100,25-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{401}{4}-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{400}{4}\right)^2=100^2=10000\)
\(------\)
\(b,B=4x^2-9y^2-6y-1\)
\(=\left(2x\right)^2-\left[\left(3y\right)^2+2.3y.1+1\right]\)
\(=\left(2x\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
Với \(x=23;y=1\), ta được:
\(B=\left(2.23-3.1-1\right)\left(2.23+3.1+1\right)\)
\(=\left(46-4\right)\left(46+4\right)\)
\(=42.50=2100\)
a) \(=\left(x-2\right)^2\)
b) \(=\left(2x+1\right)^2\)
c) \(=\left(4x-3y\right)\left(4x+3y\right)\)
d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)
e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)
f) \(=\left(x-y\right)\left(x^2+xy+y^2\right)\)
g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)
h) \(=\left(x+2\right)^3\)
i) \(=\left(1-x\right)^3\)
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
Ta có f(x) + g(x) = 4x-2.
Cho 4x - 2 = 0 ⇒ 4x = 2 ⇒ x = 1/2. Chọn A
Có:
-2x2 - 3x3 - 5x + 5x3 - x + x2 + 4x + 3 + 4x2
=2x3 + 3x2 - 2x + 3. Chọn C
Thu gọn đa thức
a,A=2x2 +x-\(\dfrac{1}{2}\)x2+5x+3
b,B=5xy+\(\dfrac{1}{2}\)x2y-\(\dfrac{2}{3}\)xy+2x2y
a: \(A=\dfrac{3}{2}x^2+6x+3\)
b: \(B=5xy-\dfrac{2}{3}xy+\dfrac{1}{2}x^2y+2x^2y=\dfrac{5}{2}x^2y+\dfrac{13}{3}xy\)
a) \(2x^2+x-\dfrac{1}{2}x^2+5x+3\)\(\)
= \(\left(2x-\dfrac{1}{2}x^2\right)+\left(x+5x\right)+3\)
= \(\dfrac{3}{2}x^2+6x+3\)
Vậy A = \(\dfrac{3}{2}x^2+6x+3\)
\(a,A=\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\\ A_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\\ b,B=x^2-6x+9-9=\left(x-3\right)^2-9\ge9\\ B_{min}=-9\Leftrightarrow x=3\)
A+B=4x2-5x+3+3x2+2x-1=7x2-3x+2
A-B=4x2-5x+3-3x2-2x+1=x2-7x+4
câu trả lời của mình nha
Ta có : \(A+B\)hay
\(4x^2-5x+3+3x^2+2x-1=7x^2-3x+2\)
Ta có : \(A-B\)hay
\(4x^2-5x+3-3x^2-2x+1=x^2-7x+4\)