Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
A = (2x - 1)(x + 2) - 3x² + (x - 1)²
= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1
= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)
= x - 1
B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)
= x³ - 8 - x³ - x² - 9 + x²
= (x³ - x³) + (-x² + x²) + (-8 - 9)
= -17
A = (2x - 1)(x + 2) - 3x² + (x - 1)²
= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1
= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)
= x - 1
B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)
= x³ - 8 - x³ - x² - 9 + x²
= (x³ - x³) + (-x² + x²) + (-8 - 9)
= -17
kết quả đây
chúc bạn học tốt
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a: \(=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)