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AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
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}
R
5 tháng 2 2022
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
5 tháng 2 2022
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
23 tháng 2 2019
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
DT
0
\(1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\) =\(1-\left(\dfrac{1}{1.14}+\dfrac{1}{14.6}+\dfrac{1}{6.34}+\dfrac{1}{34.11}\right)\)
= \(1-\left(\dfrac{1}{1}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{11}\right)\)
= 1 - \(\left(1-\dfrac{1}{11}\right)\) = \(1-\dfrac{10}{11}=\dfrac{1}{11}\)
\(=1-\dfrac{5}{2\cdot7}-\dfrac{5}{7\cdot12}-\dfrac{5}{12\cdot17}-\dfrac{5}{17\cdot22}\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=5-\dfrac{11-1}{22}=5-\dfrac{10}{22}=5-\dfrac{5}{11}=\dfrac{50}{11}\)