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Bài 1 :
(a^2+b^2)(x^2+y^2)=(ax+by)^2
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2
<=> a^2y^2 + b^2x^2 = 2abxy
<=> a^2y^2 + b^2x^2 - 2abxy = 0
<=> (ay - bx)^2 = 0
=> ay - bx = 0
=> ay = bx
=> a/x = b/y ( x,y khác 0)
Em làm thử nếu sai thì thôi ạ (vì mới học lớp 6)
a)
Ta có:
\(\left(a+b\right)^2-\left(a-b\right)^2=a^2.b^2-a^2:b^2\)
\(=a^2.b^2-a^2.\frac{1}{b^2}=a^2.\left(b^2-\frac{1}{b^2}\right)\)
Chắc thế ạ, em chỉ làm 1 phần vì sợ sai
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
ChươngII *Dạng toán rútg gọn phân thức
Bài 1.Rút gọn phân thức
a. \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}=\dfrac{-3x\left(x-1\right)}{2\left(x-1\right)}=-\dfrac{3x}{2}\)
b.\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x.2xy^2}{4y^3.2xy^2}=\dfrac{3x}{4y^3}\)
c.\(\dfrac{23\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\dfrac{23\left(x-z\right)}{6}\)
Bài 2 rút gọn các phân thức sau:
a.\(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\dfrac{x+4}{x}\)(x khác 0,x khác 4)
b.\(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
( x \(\ne-3\) )
c.\(\dfrac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+y\right)}{y}\) (y+(x+y) khác 0)
d. \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)
(x khác y)
e.\(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
(x khác -y)
f.\(\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)(x khác y,y khác 0)
g.\(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
\ (b khác 0,x khác +-1)
h. \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4x}{5x^2}\)
(x khác 0,x khác y)
i.\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
(x+y+z khác 0)
k.\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3\right)^2+2x^3y^3+\left(y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
(x khác 0,x khác +-y)
a ) \(VT=\left(x+y+z\right)^2-\left(x-y-z\right)^2\)
\(=\left(x+y+z-x+y+z\right)\left(x+y+z+x-y-z\right)\)
\(=4x\left(y+z\right)=VP\)
b ) \(VT=\left(2a+b\right)^2-\left(a+b\right)^2-3a^2\)
\(=\left(2a+b-a-b\right)\left(2a+b+a+b\right)-3a^2\)
\(=a\left(3a+2b\right)-3a^2\)
\(=3a^2+2ab-3a^2=2ab=VP\)
a) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=4x\left(y+z\right)\)
\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-\left(x^2+y^2+z^2-2xy-2xz+2yz\right)=4x\left(y+z\right)\)\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-x^2-y^2-z^2+2xy+2xz-2yz=4x\left(y+z\right)\)\(\Leftrightarrow4xy+4xz=4x\left(y+z\right)\)
\(\Leftrightarrow4x\left(y+z\right)=4x\left(y+z\right)\).
b) \(\left(2a+b\right)^2-\left(a+b\right)^2-3a^2=2ab\)
\(\Rightarrow\left(2a\right)^2+2.2a.b+b^2-\left(a^2+2ab+b^2\right)-3a^2=2ab\)
\(\Rightarrow4a^2+4ab+b^2-a^2-2ab-b^2-3a^2=2ab\)
\(\Leftrightarrow2ab=2ab\)
