a^3-a^2b-ab^2+b^3= (a+b)^3

tại a= 5,75 và b=4,25

=>(5,75+4,25)^3=10^3=1000

nguyễn hà linh Sai rồi -.-

a)   \(13,5\times5,8-8,3\times4,2-5,8\times8,3+4,2\times13,5\)

\(=13,5\times\left(5,8+4,2\right)-8,3\times\left(4,2+5,8\right)\)

\(=10\times\left(13,5-8,3\right)=10\times5,2=52\)

b) \(7,8\times55,1+92,2\times55,1-7,8\times5,1-92,2\times5,1\)

\(=55,1\times\left(7,8+99,2\right)-5,1\times\left(7,8+99,2\right)\)

\(=100\times\left(55,1-5,1\right)=100\times50=5000\)

c)    \(N=a^3-a^2b-ab^2+b^3=\left(a-b\right)^2\left(a+b\right)\)

\(=\left(5,75-4,25\right)^2.\left(5,75+4,25\right)=1,5^2\times10=22,5\)

\(M=a^3-a^2b-ab^2\)

\(M=a\left(a^2-ab-b^2\right)\)

\(M=a\left[\left(a-b\right)\left(a+b\right)-ab\right]\)

Thay a = 5,75 và b = 4,25 vào M ta được

\(M=5,75\left[\left(5,75-4,25\right)\left(5,75+4,25\right)-5,75.4,25\right]\)

\(M=5,75\left(1,5.10-24,4375\right)\)

\(M=5,75\left(15-24,4375\right)\)

\(M=5,75.\left(-9,4375\right)\)

\(M=-54,265625\)

Bài 1.

Ta có: $x+y+z=0$

$\Rightarrow x+y=-z,\ y+z=-x,\ x+z=-y$.

Suy ra: $N=(x+y)(y+z)(x+z)$$=(-z)(-x)(-y)$$=-xyz$.

Mà: $xyz=2$.

Nên: $N=-2$.

Vậy: $N=-2$.

Bài 2.

$\frac{a}{b}=\frac{10}{3}\Rightarrow a=10k,\; b=3k$

$\frac{3a-2b}{a-3b}=\frac{3\cdot10k-2\cdot3k}{10k-3\cdot3k}$

$=\frac{30k-6k}{10k-9k}$

$=\frac{24k}{k}$

$=24$

a: A=y(x-4)-5(x-4)

=(x-4)(y-5)

Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5

b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)

=0,2*10=2

d: Khi x=5,75 và y=4,25 thì

D=5,75^3-5,75^2*4,25+4,25^3

=8087/64

bn làm ơn giải chi tiết đi vs ạ

\(A=\frac{9a^5-ab^4-18a^4b+2b^5}{3a^2b^2+ab^4-6a^2b^3-2b^5}\)

\(=\frac{a\left(9a^4-b^4\right)-2b\left(9a^4-b^4\right)}{ab^2\left(3a^2+b^2\right)-2b^3\left(3a^2+b^2\right)}\)

\(=\frac{\left(9a^4-b^4\right)\left(a-2b\right)}{\left(3a^2+b^2\right)\left(ab^2-2b^3\right)}\)

\(=\frac{\left(3a^2-b^2\right)\left(3a^2+b^2\right)\left(a-2b\right)}{\left(3a^2+b^2\right)b^2\left(a-2b\right)}\)

\(=\frac{3a^2-b^2}{b^2}\)

\(=3.\left(\frac{a}{b}\right)^2-1=3.\left(\frac{2}{3}\right)^2-1=\frac{1}{3}\)

a: A=yx-4y-5x+20

=y(x-4)-5(x-4)

=(x-4)(y-5)

Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5

b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)

=0,2*10=2

d: Khi x=5,75 và y=4,25 thì

D=5,75^3-5,75^2*4,25+4,25^3

=8087/64

c: \(D=xyz-xy-yz-xz+x+y+z-1\)

=xy(z-1)-yz+y-xz+z+x-1

=xy(z-1)-y(z-1)-z(x-1)+(x-1)

=(z-1)(xy-y)-(x-1)(z-1)

=(z-1)(xy-y-1)

=(11-1)(9*10-10-1)

=10*79=790

a: \(\frac{x}{y}=\frac13\)

=>y=3x

\(\frac{14x+5y}{3x-11y}=\frac{11x+5\cdot3x}{3x-11\cdot3x}=\frac{26x}{3x-33x}=\frac{26x}{-30x}=\frac{-26}{30}=-\frac{13}{15}\)

b: \(\frac{a}{b}=\frac12\)

=>b=2a

\(\frac{11a^4-3ab^3+15a^3b+7b^4}{3a^2b^2+ab^3-6a^3b-2b^4}\)

=\(\frac{11a^4-3a\cdot\left(2a\right)^3+15a^3\cdot2a+7\left(2a\right)^4}{3a^2\cdot\left(2a\right)^2+a\cdot\left(2a\right)^3-6a^3\cdot2a-2\cdot\left(2a\right)^4}\)

\(=\frac{11a^4-3a\cdot8a^3+30a^4+7\cdot16a^4}{3a^2\cdot4a^2+a\cdot8a^3-6a^3\cdot2a-2\cdot16a^4}\)

\(=\frac{11a^4-24a^4+30a^4+112a^4}{12a^4+8a^4-12a^4-32a^4}=\frac{11-24+30+112}{12+8-12-32}=\frac{129}{-24}=\frac{-43}{8}\)