Bài 1:
a) (x+y)2=92=81
=> x2+2xy+y2=81
=> x2+2.14+y2=81
=> x2+y2=53
=> x2-2xy+y2=81-2.14=25
=> (x-y)2=25
=> x-y=5 hoặc x-y=-5
b) Câu a đã tính được x2+y2=53
c) Ta có: x3+y3=(x+y)(x2-xy+y2)=9(53-14)=9.39=351
Bài 2:
Ta có: \(x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1\)
Mà x+y=1
\(\Rightarrow1^2-4.1+1=-2\)
Bài 3:
Ta có: (x+y)3=x3+3x2y+3xy2+y3
= x3+y3+3xy(x+y)
Mà x+y=1 => (x+y)3=x3+y3+3xy=13=1
Bài 4:
Ta có: \(\left(x+y\right)^2=4^2=16\)
\(\Rightarrow x^2+2xy+y^2=16\Rightarrow10+2xy=16\)
\(\Rightarrow2xy=6\Rightarrow xy=3\)
Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4.\left(10-3\right)\)
\(=4.7=28\)
Bài 5:
Ta có: \(x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=1\left(x^2+xy+y^2\right)-3xy=x^2+xy+y^2-3xy\)
\(=x^2-2xy+y^2=\left(x-y\right)^2=1\)
Mấy bài này đầu hè làm hết rồi:))
Bài 1:
a) \(xy=14\Rightarrow x=\frac{14}{y}\)
Thay vào: \(\frac{14}{y}+y=9\)
\(\Leftrightarrow y^2+14-9y=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)
+ Nếu: \(\hept{\begin{cases}x=7\\y=2\end{cases}}\Rightarrow x-y=5\)
+ Nếu: \(\hept{\begin{cases}x=2\\y=7\end{cases}}\Rightarrow x-y=-5\)
b) Ta có: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^2=81\)
\(\Leftrightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=81-2xy=81-2.14=53\)
c) Ta có: \(x+y=9\)
\(\Leftrightarrow\left(x+y\right)^3=9^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)
\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)
Bài 1:
a,A=x-y
A2=(x-y)2
=(x2+2xy+y2)-4xy
=(x+y)2-4xy
=92-4.14 =25
b,x2+y2
=(x+y)2-2xy
=92-2.12
=53
c,x3+y3
=(x+y)3-3xy(x+y)
=93-3.14.9=351
