nhầm chứ : thay x=2 ; y=0

Sai đề khi x=2 ; y=-1 thì sai vcl

a) (2x+3)(4x²-6x+9)-2(4x³-1)+(8x-1)=15

<=>8x³+27-8x³+2+8x-1=15

<=>8x+28=15

<=>8x=-13

<=>x=-13/8

 b) (x+3)3-(x+9)(x2+27)-(5x-216) = 3x-4

<=>x³+9x²+27x+27-x³-27x-9x²-243-5x+216=3x-4

<=>-5x=3x-4

<=>8x=4

<=>x=1/2

Cộng 2 vế (1) với (2) ta được:

2P=2x²+5y²-3xy+x²-y²+2xy

=2x²+x²+5y²-y²-3xy+2xy

=3x²+4y²-xy

=>P=\(\frac{3x^2+4y^2-xy}{2}=\frac{3}{2}x^2+2y^2-\frac{1}{2}xy\)

Ta lại có: P- Q = x2 - y2 + 2xy .     

=>Q=P-(x²-y²+2xy)

=\(\frac{3}{2}x^2+2y^2-\frac{1}{2}xy-x^2+y^2-2xy=\frac{3}{2}x^2-x^2+2y^2+y^2-\frac{1}{2}xy-2xy\)

\(=\frac{1}{2}x^2+3y^2-\frac{5}{2}xy\)

x.1 +x.2+x.3+...+x.99=495             

=>x.(1+2+3+...+99)=495

=>x.\(\frac{\left(1+99\right).99}{2}=495\)

 =>x.4950=495

=>x=\(\frac{495}{4950}\)

=>x=1/10

C=100^2-99^2+98^2-97^2+......+2^2-1^2

=(100-99).(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+98+97+...+2+1

=\(\frac{\left(100+1\right).100}{2}=5050\)

\(\left(\frac{41}{371}+\frac{25}{543}.\frac{7}{11}\right).\left(1+\frac{3}{7}-\frac{20}{14}\right)=\left(\frac{41}{371}+\frac{25}{543}.\frac{7}{11}\right).\left(\frac{14}{14}+\frac{6}{14}-\frac{20}{14}\right)\)

\(=\left(\frac{41}{371}+\frac{25}{543}.\frac{7}{11}\right).\frac{0}{14}=0\)

nhìu quá @@

a, 3x³-3x²+5x+11=0

<=>3x³+3x²-6x³-6x+11x+11=0

<=>3x².(x+1)-6x.(x+1)+11.(x+1)=0

<=>(x+1)(3x²-6x+11)=0

=>x+1=0 hoặc 3x²-6x+11=0

*x+1=0 <=> x=-1

*3x²-6x+11=0

<=>2x²+x²-6x+9+2=0

<=>2x²+(x-3)²+2=0 (vô lí)

Vậy tập nghiêm của PT là S={-1}

b, 2x³-x²+3x-4=0

<=>2x³-2x²+x²-x+4x-4=0

<=>2x².(x-1)+x.(x-1)+4.(x-1)=0

<=>(x-1)(2x²+x+4)=0

<=>x-1=0 hoặc 2x²+x+4=0

*x-1=0 <=>x=1

*2x²+x+4=0

<=>x²+x²+x+1+3 = 0 ( vô lí vì \(x^2+x+1>0\)(bình phương thiếu) )

Vậy tập nghiệm của PT là S={1}

c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)

\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)

f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)

\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)