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a) Ta có:
(x2 – 2x + 5) . (x – 2)
= x2 . (x – 2) – 2x . (x – 2) + 5. (x – 2)
= x2 . x + x2 . (-2) – [2x. x + 2x.(-2) ] + 5.x + 5. (-2)
= x3 – 2x2 – (2x2 – 4x) +5x – 10
= x3 – 2x2 – 2x2 + 4x +5x – 10
= x3 +(– 2x2 – 2x2 )+ (4x +5x) – 10
= x3 – 4x2 + 9x – 10
b) Vì (x2 – 2x + 5) . (2– x) = (x2 – 2x + 5) . [-(x– 2)] = - (x2 – 2x + 5) . (x – 2)
Do đó, (x2 – 2x + 5) . (2– x) = - (x3 – 4x2 + 9x – 10) = -x3 + 4x2 - 9x + 10
a: P(1)=2+1-1=2
P(1/4)=2*1/16+1/4-1=-5/8
b: P(1)=1^2-3*1+2=0
=>x=1 là nghiệm của P(x)
P(2)=2^2-3*2+2=0
=>x=2 là nghiệm của P(x)
\(1.\)
\(\left|-0,75\right|+\frac{1}{4}-2\frac{1}{2}\)
\(=0,75+\frac{1}{4}-\frac{5}{2}\)
\(=\frac{3}{4}+\frac{1}{4}-\frac{10}{4}\)
\(=\frac{4}{4}-\frac{10}{4}\)
\(=\frac{-6}{4}=\frac{-3}{2}\)
\(2.\)
\(a,3\frac{1}{2}-\frac{1}{2}x=\frac{2}{3}\)
\(\frac{7}{2}-\frac{1}{2}x=\frac{2}{3}\)
\(\frac{1}{2}x=\frac{7}{2}-\frac{2}{3}\)
\(\frac{1}{2}x=\frac{17}{6}\)
\(x=\frac{17}{6}:\frac{1}{2}\)
\(x=\frac{17}{3}\)
Vậy x = \(\frac{17}{3}\)
\(b,3,2x+\left(-1,2\right)x+2,7\)\(=-4,9\)
\(x\cdot\left[3,2++\left(-1,2\right)\right]+2,7=-4,9\)
\(x\cdot2+2,7=-4,9\)
\(x\cdot2=-4,9-2,7\)
\(x\cdot2=-7,6\)
\(x=-7,6:2\)
\(x=-3,8\)
Vậy x=-3,8
\(3.\)
\(Có:y=f\left(x\right)\)\(=2x+\frac{1}{2}\)
\(\Rightarrow f\left(0\right)=2\cdot0+\frac{1}{2}\)\(=0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow f\left(1\right)=2\cdot1+\frac{1}{2}=2+\frac{1}{2}=\frac{4}{2}+\frac{1}{2}=\frac{5}{2}\)
\(\Rightarrow f\left(\frac{1}{2}\right)=2\cdot\frac{1}{2}+\frac{1}{2}\)\(=\frac{2}{2}+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow f\left(-2\right)=2\cdot\left(-2\right)+\frac{1}{2}=-4+\frac{1}{2}=\frac{-8}{2}+\frac{1}{2}=\frac{-7}{2}\)
`1,`
`a,`
`3x(x^2-3x+2)`
`= 3x*x^2+3x*(-3x)+3x*2`
`= 3x^3-9x^2+6x`
`b,`
`(3x^2+x) \div 3x`
`= 3x^2 \div 3x + x \div 3x`
`= x + 1/3`
a. M(x) + N(x) = 3x3 - 3x + x2 + 5 + 2x2 - x + 3x3 + 9
= (3x3 + 3x3) + ( x2 + 2x2 ) + ( -3x - x ) + (5 + 9)
= 6x3 + 3x2 - 4x + 14
b. M(x) + N(x) - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - ( 6x3 + 3x2 + 2x) = P(x)
=> 6x3 + 3x2 - 4x + 14 - 6x3 - 3x2 - 2x = P(x)
=> (6x3 - 6x3 ) + (3x2 - 3x2 ) + (-4x - 2x ) + 14 = P(x)
=> -6x + 14 = P(x)
Ta có : -6x + 14 = 0
=> -6x = -14
=> x = 7/3
=> Đa thức P(x) = -6x + 14 có nghiệm là 7/3
=>
\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{\left(5.5.4\right)^4}{\left(5.5.4\right)^5}=\frac{1^4}{1^5}=1\)
bài 1: f(x) + 2f(2-x)=3x (1)
f(2-x)+2[(2-(2-x)]=3(2-x) suy ra f(2-x)+2f(x)=6-3x suy ra 2f(2-x)+4f(x)=12-6x (2)
Lấy (2)-(1) ta có: 4f(x)-f(x)=12-6x-3x suy ra f(x)=4-3x
vậy f(2)=4-3*2=-2
Bài 2 tương tự: f(x)+3f(1/x)=x^2 (1)
f(1/x)+3f(x)=1/x^2 suy ra 3f(1/x)+9f(x)=3/x^2 (2)
Lấy (2)-(1) ta có: 9f(x)-f(x)=3/x^2-x^2 suy ra f(x)=(3-x^4)/8x^2
Vậy f(2)=(3-2^4)(8*2^2)=-13/32
a: (x^2-3x+2)(x-2)
=x^3-2x^2-3x^2+6x+2x-4
=x^3-5x^2+8x-4
b: (x^2-3x+2)(2-x)
=-(x-2)(x^2-3x+2)
=-(x^3-5x^2+8x-4)
=-x^3+5x^2-8x+4