chứng minh các phân thức sau
a) \(\frac{3y}{4}=\frac{6xy}{8x}\left(x\ne0\right)\)
b)\(\frac{-3x^2}{2y}=\frac{3x^2}{-2y}\left(y\ne0\right)\)
c)\(\frac{2\left(x-y\right)}{3\left(y-x\right)}=\frac{-2}{3}\left(x\ne y\right)\)
a, Ta có : \(\frac{3y}{4}=\frac{3y}{4}.1=\frac{3y}{4}.\frac{2x}{2x}=\frac{6xy}{8x}\) ( đpcm )
b, Ta có : \(6x^2y=6x^2y\)
=> \(3x^2.2y=\left(-3x^2\right).\left(-2y\right)\)
=> \(\frac{-3x^2}{2y}=\frac{3x^2}{-2y}\) ( đpcm )
c, Ta có : \(6x-6y=6x-6y\)
=> \(6x-6y=-6y+6x\)
=> \(6\left(x-y\right)=-6\left(y-x\right)\)
=> \(2\left(x-y\right).3=-2\left(y-x\right).3\)
=> \(\frac{2\left(x-y\right)}{3\left(y-x\right)}=\frac{-2}{3}\) ( đpcm )
bài 2 : rút gọn các phân thức sau :
a.\(\frac{x^2-16}{4x-x^2}\left(x\ne0,x\ne4\right)\)
b.\(\frac{x^2+4x+3}{2x+6}\left(x\ne-3\right)\)
c.\(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\left(y\ne0;x+y\ne0\right)\)
d. \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\left(x\ne y\right)\)
e. \(\frac{x^2-xy}{3xy-3y^2}\left(x\ne y,y\ne0\right)\)
f. \(\frac{4x^2-4xy}{5x^3-5x^2y}\left(x\ne0,x\ne y\right)\)
g. \(\frac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\)
https://hoc24.vn/hoi-dap/question/697806.html
cmr nếu\(a\left(z+y\right)=b\left(z+x\right)=c\left(x+y\right);a\ne b\ne c\ne0\Rightarrow\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
đề đúng mà bn
đề đúng thì giải giùm ik bạn ơi
rút gọn các phân thức
a) \(\frac{x^2-16}{4x-x^2}\left(x\ne0,x\ne4\right)\) d) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ab}\)
b) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\left(x\ne y\right)\) c) \(\frac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\)
e)\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}\)
Lời giải:
a) \(\frac{x^2-16}{4x-x^2}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
c)
\(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)
d)
Biểu thức không rút gọn được
e)
\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b)^3-3ab(a+b)+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ac-bc+2ab)-3ab(a+b+c)+3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc}{a^2+b^2+c^2-ab-bc-ac}=a+b+c+\frac{3abc}{a^2+b^2+c^2-ab-bc-ac}\)
1. Cho \(a,b\in Z;a,b\ne0;a\ne3b;a\ne-5b\). C/m giá trị A là 1 số nguyên lẻ \(A=\frac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\frac{a^2b+5ab^2}{a^2-3ab}\)
2. Cho \(x+y+z=1\)và \(x\ne-y;y\ne-z;z\ne-x\)
Tính giá trị biểu thức \(Q=\frac{xy+z}{\left(x+y\right)^2}.\frac{yz+x}{\left(y+z\right)^2}.\frac{zx+y}{\left(z+x\right)^2}\)
3. Cho \(xyz=1\).Tính \(P=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(z+\frac{1}{z}\right)^2-\left(x+\frac{1}{x}\right)\left(y-\frac{1}{y}\right)\left(z-\frac{1}{z}\right)\)
1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa
CMR: Nếu a(y+z)=b(z+x)=c(x+y)\(\left(a\ne b\ne c\ne0\right)\)thì \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
Chmr nếu:
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}vớix\ne y,yz\ne1,xz\ne1,x\ne0,y\ne0,z\ne0\)
thì: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Rút gọn: \(\frac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\frac{x\left(y^2+z\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(x\ne y\ne z\ne0\)
Rút gọn phân thức P=\(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\) với \(x\ne0,y\ne0,x\ne-y\)
Với đk trên ta có:
P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{y}{x+y}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x-y}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\frac{x-y}{xy}.\left(xy-\left(x+y\right)^2\right).\frac{1}{x^2+xy+y^2}\)
\(=\frac{2}{x}+\frac{x-y}{xy}\)
\(=\frac{x+y}{xy}\)
