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AH
Akai Haruma
Giáo viên
7 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\neq 0;-1$
\(=\left(\frac{2x^2+3x}{(x+1)(x^2-x+1)}+\frac{x+1}{(x+1)(x^2-x+1)}\right).\frac{x^2-x+1}{x}\)
\(=\frac{2x^2+3x+x+1}{(x+1)(x^2-x+1)}.\frac{x^2-x+1}{x}=\frac{2x^2+4x+1}{x(x+1)}\)
b. ĐKXĐ: $x\neq 0; 1;2$
\(=\frac{x-(x-1)}{x(x-1)}:\frac{(x+1)(x-1)-(x-2)(x+2)}{(x-2)(x-1)}=\frac{1}{x(x-1)}:\frac{3}{(x-2)(x-1)}\)
\(=\frac{1}{x(x-1)}.\frac{(x-2)(x-1)}{3}=\frac{x-2}{3x}\)
c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)
11 tháng 7 2023
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
20 tháng 12 2022
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
MH
31 tháng 1 2021
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.
31 tháng 1 2021
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=0
hay x=1(nhận)
Vậy: S={1}
AD
23 tháng 7 2023
\(a,=\left(\dfrac{1-x}{x}+\dfrac{x^3-x}{x}\right)\times\dfrac{x}{x-1}\\ =\dfrac{1-x+x^3-x}{x}\times\dfrac{x}{x-1}\\ =\dfrac{1-2x+x^3}{x-1}\\ b,=\left(\dfrac{x-x^2}{x.x^2}\right).\dfrac{x^2}{y}+\dfrac{x}{y}\\ =\dfrac{x-x^2}{xy}+\dfrac{x}{y}\\ =\dfrac{x-x^2+x^2}{xy}=\dfrac{x}{xy}=\dfrac{1}{y}\)
\(c,=\dfrac{3}{x}-\dfrac{2}{x}\times x+\dfrac{x}{3}\\ =\dfrac{3}{x}-2+\dfrac{x}{3}\\ =\dfrac{3-2x+x^2}{3x}\)
TL
19 tháng 3 2021
1. \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x}-\dfrac{2x}{x^2+x}=\dfrac{3}{x^2+x}\)
\(\Rightarrow x+1-2x=3\)
\(\Leftrightarrow1-x=3\)
\(\Leftrightarrow-x=2\\ \Leftrightarrow x=-2\)
Vậy phương trình có nghiệm duy nhất \(x=-2\)
2. \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2+2x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2=2\\ \Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0 \)
\(\Leftrightarrow x=0\) hoặc x + 1= 0
⇔ x = 0 hoặc x= -1
Vậy phương trình có tập nghiệm là S={0;-1}
19 tháng 3 2021
1) ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\)
\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{2x}{x\left(x+1\right)}=\dfrac{3}{x\left(x+1\right)}\)
Suy ra: \(x+1-2x=3\)
\(\Leftrightarrow-x+1=3\)
\(\Leftrightarrow-x=2\)
hay x=-2(thỏa ĐK)
Vậy: S={-2}
5 tháng 1
d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)
=>\(x+3-6\left(x-2\right)=-5\)
=>x+3-6x+12=-5
=>-5x+15=-5
=>-5x=-20
=>x=4(nhận)
e: ĐKXĐ: x<>-2
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)
=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)
=>\(2x^2-4x+8-2x^2-16=5x+10\)
=>5x+10=-4x-8
=>9x=-18
=>x=-2(loại)
f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)
\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)
=>\(\left(x^2-1\right)\cdot\left(x^3+1-x^3+1\right)=2\left(x^2+4x+4\right)\)
=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)
=>8x=-10
=>x=-5/4(nhận)
14 tháng 12 2021
\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
\(c,=\dfrac{3x^2+2x+1+x^2-2x+1-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)