Câu hỏi của Hoài An

Question

1) $\dfrac{4x+7}{x-1}$ = $\dfrac{12x+5}{3x+4}$2) $\dfrac{x}{x-1}$ - $\dfrac{2x}{x^{2^{ }}-1}$ = 03) $\dfrac{1}{3-x}$ - $\dfrac{14}{x^2-9}$ = 14) $\dfrac{x+1}{x-1}$ - $\dfrac{x-1}{x+1}$...

Minh Hồng · Accepted Answer

1/ $\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}$ (1)Điều kiện: $\left\{{}\begin{matrix}x-1\ne0\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\x\ne-\dfrac{4}{3}\end{matrix}\right.$(1) $\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\
\Leftrightarrow\left(16+21+12-5\right)x=-5-28\
\Leftrightarrow44x=-33\
\Leftrightarrow x=-\dfrac{3}{4}$ (Thỏa mãn)Vậy $x=-\dfrac{3}{4}$.2/ $\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0$ (2)Điều kiện: $x\ne\pm1$(2)$\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\
\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\
\Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\
\Leftrightarrow x\left(x+1\right)-2x=0\
\Leftrightarrow x^2+x-2x=0\
\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$So sánh với điều kiện $\Rightarrow x=0$ là nghiệm của PT.3/ $\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1$ (3)Điều kiện: $x\ne\pm3$(3)$\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\
\Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\
\Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\
\Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0$ (Vô nghiệm do $x^2+x+8>0\qquad\forall x$).Vậy PT vô nghiệm.4/ $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}$ (4)Điều kiện: $x\ne\pm1$(4)$\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\
\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\
\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\
\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1$ (loại)Vậy PT vô nghiệm.5/ $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$ (5)Điều kiện: $x\ne0$(5)$\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2$Đặt $t=x+\dfrac{1}{x}$, ta có: $t=t^2-2\
\Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\
\Leftrightarrow\left[{}\begin{matrix}t=2\t=-1\end{matrix}\right.$Với $t=2$ ta có: $x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1$ (thỏa mãn)Với $t=-1$ ta có: $x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0$ (vô nghiệm).Vậy $x=1$ là nghiệm PT.6/ $\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}$ (6)Điều kiện: $x\ne-1$(6)$\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\
\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x-1=0\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.$$x-1=0\Leftrightarrow x=1$ (Thỏa mãn)$\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\
\Leftrightarrow x^2-x+3=0$ (vô nghiệm).Vậy $x=1$ là nghiệm PT. Câu trả lời: 1/ $\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}$ (1)Điều kiện: $\left\{{}\begin{matrix}x-1\ne0\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\x\ne-\dfrac{4}{3}\end{matrix}\right.$(1) $\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\
\Leftrightarrow\left(16+21+12-5\right)x=-5-28\
\Leftrightarrow44x=-33\
\Leftrightarrow x=-\dfrac{3}{4}$ (Thỏa mãn)Vậy $x=-\dfrac{3}{4}$.2/ $\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0$ (2)Điều kiện: $x\ne\pm1$(2)$\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\
\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\
\Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\
\Leftrightarrow x\left(x+1\right)-2x=0\
\Leftrightarrow x^2+x-2x=0\
\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$So sánh với điều kiện $\Rightarrow x=0$ là nghiệm của PT.3/ $\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1$ (3)Điều kiện: $x\ne\pm3$(3)$\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\
\Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\
\Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\
\Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0$ (Vô nghiệm do $x^2+x+8>0\qquad\forall x$).Vậy PT vô nghiệm.4/ $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}$ (4)Điều kiện: $x\ne\pm1$(4)$\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\
\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\
\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\
\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1$ (loại)Vậy PT vô nghiệm.5/ $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$ (5)Điều kiện: $x\ne0$(5)$\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2$Đặt $t=x+\dfrac{1}{x}$, ta có: $t=t^2-2\
\Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\
\Leftrightarrow\left[{}\begin{matrix}t=2\t=-1\end{matrix}\right.$Với $t=2$ ta có: $x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1$ (thỏa mãn)Với $t=-1$ ta có: $x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0$ (vô nghiệm).Vậy $x=1$ là nghiệm PT.6/ $\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}$ (6)Điều kiện: $x\ne-1$(6)$\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\
\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x-1=0\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.$$x-1=0\Leftrightarrow x=1$ (Thỏa mãn)$\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\
\Leftrightarrow x^2-x+3=0$ (vô nghiệm).Vậy $x=1$ là nghiệm PT.

Nguyễn Lê Phước Thịnh · Answer

1) ĐKXĐ: $x\notin\left\{1;-\dfrac{4}{3}\right\}$Ta có: $\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}$$\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)$$\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5$$\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0$$\Leftrightarrow20x+33=0$$\Leftrightarrow20x=-33$$\Leftrightarrow x=-\dfrac{33}{20}$(nhận)Vậy: $S=\left\{-\dfrac{33}{20}\right\}$2) ĐKXĐ: $x\notin\left\{1;-1\right\}$Ta có: $\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0$$\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0$Suy ra: $x^2+x-2x=0$$\Leftrightarrow x^2-x=0$$\Leftrightarrow x\left(x-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\x=1\left(loại\right)\end{matrix}\right.$Vậy: S={0}3) ĐKXĐ: $x\notin\left\{3;-3\right\}$Ta có: $\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1$$\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1$$\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}$Suy ra: $-x-3-14=x^2-9$$\Leftrightarrow x^2-9=-x-17$$\Leftrightarrow x^2-9+x+17=0$$\Leftrightarrow x^2+x+8=0$$\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0$$\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0$(vô lý)Vậy: $S=\varnothing$4) ĐKXĐ: $x\notin\left\{1;-1\right\}$Ta có: $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}$$\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}$Suy ra: $x^2+2x+1-\left(x^2-2x+1\right)=4$$\Leftrightarrow x^2+2x+1-x^2+2x-1=4$$\Leftrightarrow4x=4$hay x=1(loại)Vậy: $S=\varnothing$5) ĐKXĐ: $x\ne0$Ta có: $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$$\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}$$\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)$$\Leftrightarrow x^4+x^2=x^5+x$$\Leftrightarrow x^5+x-x^4-x^2=0$$\Leftrightarrow x\left(x^4-x^3-x+1\right)=0$$\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0$$\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0$$\Leftrightarrow x\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0$mà $x^2+x+1>0$nên $x\cdot\left(x-1\right)^2=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x-1=0\end{matrix}\right.\Leftrightarrow x=1$Vậy: S={1}6) ĐKXĐ: $x\in R$Ta có: $\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}$$\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)$$\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0$$\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0$$\Leftrightarrow\left(x-1\right)\left(-x^2+x-3\right)=0$$\Leftrightarrow-\left(x-1\right)\left(x^2-x+3\right)=0$mà $x^2-x+3>0$nên x-1=0hay x=1(nhận)Vậy: S={1}Câu trả lời: 1) ĐKXĐ: $x\notin\left\{1;-\dfrac{4}{3}\right\}$Ta có: $\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}$$\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)$$\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5$$\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0$$\Leftrightarrow20x+33=0$$\Leftrightarrow20x=-33$$\Leftrightarrow x=-\dfrac{33}{20}$(nhận)Vậy: $S=\left\{-\dfrac{33}{20}\right\}$2) ĐKXĐ: $x\notin\left\{1;-1\right\}$Ta có: $\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0$$\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0$Suy ra: $x^2+x-2x=0$$\Leftrightarrow x^2-x=0$$\Leftrightarrow x\left(x-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\x=1\left(loại\right)\end{matrix}\right.$Vậy: S={0}3) ĐKXĐ: $x\notin\left\{3;-3\right\}$Ta có: $\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1$$\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1$$\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}$Suy ra: $-x-3-14=x^2-9$$\Leftrightarrow x^2-9=-x-17$$\Leftrightarrow x^2-9+x+17=0$$\Leftrightarrow x^2+x+8=0$$\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0$$\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0$(vô lý)Vậy: $S=\varnothing$4) ĐKXĐ: $x\notin\left\{1;-1\right\}$Ta có: $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}$$\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}$Suy ra: $x^2+2x+1-\left(x^2-2x+1\right)=4$$\Leftrightarrow x^2+2x+1-x^2+2x-1=4$$\Leftrightarrow4x=4$hay x=1(loại)Vậy: $S=\varnothing$5) ĐKXĐ: $x\ne0$Ta có: $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$$\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}$$\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)$$\Leftrightarrow x^4+x^2=x^5+x$$\Leftrightarrow x^5+x-x^4-x^2=0$$\Leftrightarrow x\left(x^4-x^3-x+1\right)=0$$\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0$$\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0$$\Leftrightarrow x\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0$mà $x^2+x+1>0$nên $x\cdot\left(x-1\right)^2=0$$\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x-1=0\end{matrix}\right.\Leftrightarrow x=1$Vậy: S={1}6) ĐKXĐ: $x\in R$Ta có: $\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}$$\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)$$\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0$$\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0$$\Leftrightarrow\left(x-1\right)\left(-x^2+x-3\right)=0$$\Leftrightarrow-\left(x-1\right)\left(x^2-x+3\right)=0$mà $x^2-x+3>0$nên x-1=0hay x=1(nhận)Vậy: S={1}

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP