Ta thấy : \(\left(x+1\right)^{2010}\ge0;\left(y-1\right)^{2000}\ge0\)

\(\Rightarrow\)\(\hept{\begin{cases}x+1=0\\y-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

Tại x = -1; y = 1, ta có :

3(-1)⁷ - 5.1⁶ + 1 = - 3 - 5 +1 = - 7

TLMJFDLIIS HFIEHFU ưAUDSEIq

1, Tính giá trị biểu thức sau tại x+y+1=0

\(D=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\left(1\right)\)

Ta có: x + y + 1 = 0 => x + y = -1

(1) \(\Leftrightarrow x^2.\left(-1\right)-y^2.\left(-1\right)+\left(x-y\right)\left(x+y\right)+2.\left(-1\right)+3\)

\(=y^2-x^2+\left(x-y\right)\left(-1\right)-2+3\)

\(=\left(y-x\right)\left(y+x\right)-\left(x-y\right)+1\)

\(=\left(y-x\right).\left(-1\right)-x+y+1\)

\(=-y+x-x+y+1\)

\(=1\)

2, Cho xyz=2 và x+y+z=0

Tính giá trị biểu thức

\(M=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có: x + y + z = 0

=> x + y = -z (1)

=> y + z = -x (2)

=> x + z = -y (3)

Từ (1);(2);(3) 

=> \(M=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)<=> (-z).(-x).(-y) = 0

1, x+y+z=1

\(D=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\)

\(=\left(x+y\right)\left(x^2-y^2+2\right)+x^2-y^2+3\)

\(=\left(x+y\right)\left(x^2-y^2+2\right)+\left(x^2-y^2+2\right)+1\)

\(=\left(x^2-y^2+2\right)\left(x+y+1\right)+1\)

=1 (vì x+y+1=0)

2, x+y+z=0 <=> \(\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{cases}}\)

Nhân theo vế ta được: xyz=\(-\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(\Rightarrow2=-\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

=> (x+y)(y+z)(z+x)=-2

3x²y²z² = x³y³ y³z³ z³x³ 
(3x²y²z²) / (x³y³ y³z³ z³x³) = 1
3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1
(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3
(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3
z²/(xy) x/(yz) y²/(zx) = 1/3
Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3

Ta có : 

\(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)

\(=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}\)

Do x + y + z = 0 => x+y = -z ; y+z = -x ; z+x = -y

\(\Rightarrow A=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=\frac{\left(-1\right).xyz}{xyz}=-1\)

3r3reR

Bài 1 : 

\(N=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....