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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
c) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=1\) đặt x-2 =t " cho bé hệ số lại
ĐK : \(\left\{\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t\ne0\\t\ne-2\end{matrix}\right.\)
\(\frac{t-1}{t}=\frac{t}{t-2}\Leftrightarrow\left(t-1\right)\left(t-2\right)=t^2\Leftrightarrow t^2-3t+2=t^2\Rightarrow-3t=-2\)
\(t=\frac{2}{3}\Rightarrow x=2+\frac{2}{3}=\frac{8}{3}\)
a) \(A=\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3x}\) xem lại đề thấy cái mẫu VP vô duyên thế!
b) \(B=\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\) MSC=(x^3-1)
\(B=\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(4x^2-1\right)}{MSC}=\frac{\left(2x^2+2x+2\right)+\left(2x^2+x-3\right)-4x^2+1}{MSC}=0\)
\(B=0\Leftrightarrow\frac{3x}{MSC}=0=>x=0\) thảo mãn đk x khác 1
Kết luận: x=0 là nghiệm duy nhất.
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
a/ Do \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{3}{x}-1}-\frac{2}{x+\frac{3}{x}-3}=-1\)
Đặt \(x+\frac{3}{x}-3=a\) ta được:
\(\frac{3}{a+2}-\frac{2}{a}=-1\)
\(\Leftrightarrow3a-2\left(a+2\right)=-a\left(a+2\right)\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\)
b/ Đặt \(x^2+2x+\frac{5}{2}=a>0\)
Phương trình trở thành:
\(\frac{1}{\left(a-\frac{1}{2}\right)^2}+\frac{1}{\left(a+\frac{1}{2}\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow4\left(a+\frac{1}{2}\right)^2+4\left(a-\frac{1}{2}\right)^2=5\left(a^2-\frac{1}{4}\right)^2\)
\(\Leftrightarrow8a^2+2=5\left(a^4-\frac{1}{2}a^2+\frac{1}{16}\right)\)
\(\Leftrightarrow5a^4-\frac{21}{2}a^2-\frac{27}{16}=0\Rightarrow\left[{}\begin{matrix}a^2=\frac{9}{4}\\a^2=-\frac{3}{20}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x+\frac{5}{2}=\frac{3}{2}\\x^2+2x+\frac{5}{2}=-\frac{3}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2+\frac{2x^2}{x^2-1}-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
Đặt \(\frac{2x^2}{x^2-1}=a\)
\(\Rightarrow a^2-a-\frac{10}{9}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{5}{3}\\a=-\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2x^2}{x^2-1}=\frac{5}{3}\\\frac{2x^2}{x^2-1}=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=-5\left(l\right)\\x^2=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{1}{2}\)
d/ĐKXĐ: ...
\(\Leftrightarrow\left(x^2+\frac{36}{x^2}\right)-13\left(x-\frac{6}{x}\right)=0\)
Đặt \(x-\frac{6}{x}=a\Rightarrow x+\frac{36}{x^2}=a^2+12\)
\(\Rightarrow a^2-13a+12=0\Rightarrow\left[{}\begin{matrix}a=1\\a=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=1\\x-\frac{6}{x}=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-6=0\\x^2-12x-6=0\end{matrix}\right.\)