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Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
Bài 1:
a/\(xy\ne0\), nhân cả tử và mẫu với \(xy\) ta được:
\(\frac{x^2+y^2-2xy}{x^2-y^2}=\frac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x-y}{x+y}\)
b/ \(x\ne\pm1\), nhân cả tử và mẫu với \(x^2-1=\left(x-1\right)\left(x+1\right)\) ta được:
\(\frac{x^2-1-2\left(x-1\right)}{x^2-1-\left(x^2-2\right)}=\frac{x^2-2x+1}{1}=\left(x-1\right)^2\)
c/ \(x\ne\pm1\), nhân cả tử và mẫu với \(\left(x-1\right)\left(x+1\right)\) ta được:
\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2}=\frac{x^2+2x+1-x^2+2x-1}{x^2-1-x^2+2x-1}=\frac{4x}{2x}=2\)
Bài 2:
a/ Xem lại đề, thấy có vẻ ko đối xứng lắm, \(\frac{2x+1}{2x-2}\) hay \(\frac{2x+1}{2x-1}\) bạn?
b/ \(x\ne\left\{-1;0;1\right\}\)
\(B=\left(\frac{1}{x\left(x+1\right)}+\frac{x-2}{x+1}\right):\left(\frac{x^2-2x+1}{x}\right)\)
\(B=\left(\frac{1}{x\left(x+1\right)}+\frac{x\left(x+2\right)}{x\left(x+1\right)}\right).\frac{x}{\left(x-1\right)^2}\)
\(B=\frac{\left(x^2+2x+1\right)}{x\left(x+1\right)}.\frac{x}{\left(x-1\right)^2}\)
\(B=\frac{\left(x+1\right)^2}{x\left(x+1\right)}.\frac{x}{\left(x-1\right)^2}=\frac{x+1}{\left(x-1\right)^2}\)
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
Bài 1:
a/ \(x\ne1;2\)
\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x-2-7x+7+1=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Rightarrow x=1\) (loại)
Vậy pt vô nghiệm
b/ \(x\ne\frac{3}{2}\)
\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)
\(\Leftrightarrow20x+30-15-8x+12=0\)
\(\Leftrightarrow12x+27=0\)
\(\Rightarrow x=-\frac{9}{4}\)
c/ \(x\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Rightarrow x=4\)
Bài 1:
d/\(x\ne\pm3\)
\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
\(\Rightarrow0=0\)
Vậy pt có vô số nghiệm \(x\ne\pm3\)
e/ \(x\ne\pm1\)
\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Leftrightarrow x^2-2x+1+2+3x-3=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)
a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{x+1}{x}\div\frac{x^2-1}{x}=\frac{x+1}{x}\cdot\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x-1}\)
b) \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right)\div\left(\frac{1}{x+2}+\frac{1}{x-2}\right)=\frac{\left(x-2\right)^2-\left(x+2^2\right)}{\left(x^2-4\right)^2}\div\frac{x-2+x+2}{x^2-4}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x^2-4\right)^2}\cdot\frac{x^2-4}{2x}=\frac{2x\cdot\left(-4\right)}{x^2-4}\cdot\frac{1}{2x}=\frac{-4}{x^2-4}\)
a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x^2-1}{x}}=\frac{x+1}{x}\cdot\frac{x}{x^2-1}=\frac{1}{x-1}\)
b) \(\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2^2\right)}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(\Leftrightarrow\left(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(\Leftrightarrow\left(\frac{x^2-4x+4-x^2-4x-4}{\left[\left(x-2\right)\left(x+2\right)\right]^2}\right):\left(\frac{x-2+x+2}{x^2-4}\right)\)
\(\Leftrightarrow\frac{-8x}{\left(x^2-4\right)^2}\cdot\frac{x^2-4}{2x}\)\(\Leftrightarrow-\frac{4}{x^2-4}\)
d) \(\frac{3x}{x^3-1}+\frac{x-1}{x^2+x+1}\Leftrightarrow\frac{3x}{x^3-1}+\frac{\left(x-1\right)^2}{x^3-1}\)
\(\Leftrightarrow\frac{x^2-2x+1+3x}{x^3-1}=\frac{x^2+x+1}{x^3-1}=\frac{1}{x-1}\)
còn lại chút giải tiếp !!!
Ahuhu, không ai biết cách giải ư ? T^T