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Ta có: \(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)
\(=\left[\frac{1}{\left(x+y\right)^2}-\frac{1}{\left(x+y\right)\left(x-y\right)}\right].\frac{\left(y+x\right)\left(y-x\right)}{4xy}\)
\(=\frac{1}{x+y}\left(\frac{1}{x+y}-\frac{1}{x-y}\right).\frac{\left(x+y\right)\left(y-x\right)}{4xy}\)
\(=\frac{-2y}{\left(x+y\right)\left(x-y\right)}.\frac{x-y}{-4xy}\)
\(=\frac{1}{\left(x+y\right).2x}\)
Kb với mình nha mn!
ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
a,ĐKXĐ \(x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)
b,\(\Leftrightarrow3x^2+6x+12=0\)
\(\Leftrightarrow3\left(x^2+2x+1\right)+9=0\)
\(\Leftrightarrow3\left(x+1\right)^2+9=0\)(VÔ LÝ VÌ 3(x+1)2>=0 =>3(x+1)2+9>0)
vì vây ko có giá trị x để F =0
C, VỚI ĐKXĐ trên ,ta có
\(F=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{3}{x-2}\)
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)
a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)
\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)
\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)
b, \(x=\frac{1}{2}\)
\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)
c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)
\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)
\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)
\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)
\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)
bài 1.2 làm như sau:
x3 - 3x2+3x-1=0
x3-3x2.1+3x.12-13=0
áp dụng HĐT số 5 trong sách ta có
(x-1)3=0
=> x-1=0
x=1
= ( x/(x-6)(x+6) - x-6/x(x+6) ) : 2x-6/x2 + 6x + 6/6-x
=( x2/x(x+6)(x-6) - (x -6 )(x-6)/x(x+6)(x-6) ) : .....
= (12x -36 / x(x+6)(x-6) : 2x-6/ x2 + 6x )+ 6/6-x
=6/x-6 + 6/6-x
= 6-6/ x-6
=0/x-6
câu trước mình thiếu 6/6-x
Cái này dễ mà bn
Ta có:\(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\left(ĐK:x\ne2;-3\right)\)
\(\Leftrightarrow A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)
\(\Leftrightarrow A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x+4}{x-2}\)
\(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(\Leftrightarrow A=\frac{x}{\left(2+3\right)}^2-\frac{5}{x^3-6}+\left(2-x\right)\)
\(\Leftrightarrow A=\frac{x}{5}^2-\frac{5}{x^3-6}+\left(2-x\right)\)
Ps: Không chắc đâu nhé! Thánh đây mới lớp 6 thôi