Cho x+y=2, tinh gia tri cua bieu thuc:
M=3(x^2+y^2)-(x^3+y^3)+1
Bai 2:Cho a+b=5,tinh gia tri bieu thuc:
M=3a^2-2a+3b^2-2b+6ab+100
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
Cho x - y = 11. Tinh gia tri cua bieu thuc : M = x^3-3xy(x- y) - y^3 -x^2 + 2xy - y^2
Ta có: x^3 -3xy(x-y) -y^3 -x^2 + 2xy-y^2
= x^3 -y^3 - 3xy(x-y) -( x^2 -2xy+y^2)
= (x-y)(x^2+xy +y^2) - 3xy(x-y) -(x-y)^2
= (x-y)(x^2+xy+y^2 -3xy-x+y)
=11( x^2 -2xy+y^2 -x+y)
= 11[ (x-y)^2 -(x-y)]
= 11[ 11^2 -11]
= 11^3 -11^2=...
tinh gia tri cua bieu thuc x-y=0 sao cho x.(x^2+y^2)+y.(-y^2-X^2)+5
Cho phan thuc B=(3\y+3)+(1\y-3)-(18\9-y2)
a)Tim dieu kien cua y de gia tri cua bieu thuc B duoc xac dinh
b)Rut gon bieu thuc B
c)Tinh gia tri cua B de B co gia tri nguyen
tinh gia tri cua bieu thuc; P=x^3+x^2*y-2x^2-y(x+y)+3y+x+2018với x+y=2
moi ng giup mh nhanh nhanh!
a, Cho x+y = 2 va x^2+y^2=10. Tinh gia tri bieu thu x^3 +y^3
b, Cho x+y=a va x^2+ y^2= b. Tinh x^3 +y^3 theo a,b
a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x3 + y3 = 26
a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)
vậy............
b, \(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\)
\(\Rightarrow xy=\frac{a^2-b}{2}\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)
Vậy....
tinh gia tri cua bieu thuc A=\(x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
Sửa đề: x+y=1
\(A=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\)
\(=1-3xy+3xy\left[1-2xy\right]+6x^2y^2\)
=1
1) Cho bieu thuc A=\(3+\frac{2}{x-1}\). Tinh gia tri cua bieu thuc A khi |2x-3|=1
2) Rut gon bieu thuc B=\(\frac{x}{x-1}\)-\(\frac{x-5}{x+1}\)-\(\frac{3-x}{1-x^2}\)
3) Tim cac gia tri nguyen cua x de bieu thuc \(\frac{B}{A}\)co gia tri nguyen duong
Cho x^2+y^2= 1. Tinh gia tri bieu thuc: 2(x^6+y^6)-3(x^4+y^4)?
2(x^6+y^6)-3(x^4+y^4)
= 2x^4(x² - 1) + 2y^4(y² - 1) - (x^4 + y^4)
= - 2x^4 .y² - 2y^4 .x² - [(x² +y²)² - 2x².y²]
= - 2x²y².(x² + y²) - 1 + 2x².y²
= - 2x²y² - 1 + 2x²y²
= - 1.