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\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
a/ S=\(\left(\frac{x}{\left(x+6\right)\left(x-6\right)}-\frac{x-6}{x\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2x-6}-\frac{x}{x-6}\)
S=\(\frac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}.\frac{x\left(x+6\right)}{2x-6}-\frac{x}{x-6}\)=\(\frac{\left(x-x+6\right)\left(x+x-6\right)}{\left(x-6\right)\left(2x-6\right)}-\frac{x}{x-6}\)
= \(\frac{6\left(2x-6\right)}{\left(x-6\right)\left(2x-6\right)}-\frac{x}{x-6}\)= \(\frac{6}{\left(x-6\right)}-\frac{x}{x-6}\)\(\frac{6-x}{x-6}=-1\)
b/ S luôn =-1 với mọi x
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
a)\(\frac{x^2+3x+2}{3x+6}=\frac{x^2+2x+x+2}{3\cdot\left(x+2\right)}=\frac{\left(x^2+2x\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}=\frac{x\cdot\left(x+2\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}\)
\(=\frac{\left(x+2\right)\cdot\left(x+1\right)}{3\cdot\left(x+2\right)}=\frac{x+1}{3}\)
b) \(\frac{2x^2+x-1}{6x-3}=\frac{2x^2+2x-x-1}{3\cdot\left(2x-1\right)}=\frac{\left(2x^2+2x\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}\)
\(=\frac{2x\cdot\left(x+1\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{\left(2x-1\right)\cdot\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{x+1}{3}\)
\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
1)
ĐKXĐ: x\(\ne\)3
ta có :
\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)
để biểu thức A có giá trị = 1
thì :\(\frac{x-3}{2}\)=1
=>x-3 =2
=>x=5(thoả mãn điều kiện xác định)
vậy để biểu thức A có giá trị = 1 thì x=5
1)
\(A=\frac{x^2-6x+9}{2x-6}\)
A xác định
\(\Leftrightarrow2x-6\ne0\)
\(\Leftrightarrow2x\ne6\)
\(\Leftrightarrow x\ne3\)
Để A = 1
\(\Leftrightarrow x^2-6x+9=2x-6\)
\(\Leftrightarrow x^2-6x-2x=-6-9\)
\(\Leftrightarrow x^2-8x=-15\)
\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)
a) \(\frac{x^2+2x-3}{x^2+x-6}\)
\(=\frac{x^2+3x-x-3}{x^2-2x+3x-6}\)
\(=\frac{x\left(x+3\right)-\left(x+3\right)}{x\left(x-2\right)+3\left(x-2\right)}\)
\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x-1}{x-2}\)
b) \(\frac{2x^2+2x+2}{x^2+6x-7}\)
\(=\frac{2\left(x^2+x+1\right)}{x^2+7x-x-7}\)
\(cogidosaisai?\)