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2 tháng 1 2018
\(a,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\dfrac{15x-6-8}{4}=\dfrac{7x-15\left(x-7\right)}{3}\)
\(\Leftrightarrow\dfrac{15x-14}{4}=\dfrac{7x-15x+105}{3}\)
\(\Leftrightarrow\dfrac{45x-42}{12}=\dfrac{-32x+420}{12}\)
\(\Leftrightarrow45x+32x=420+42\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
\(b,\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)
\(\Leftrightarrow\dfrac{2x+10+3-2x}{4}=\dfrac{6x-7-x}{6}\)
\(\Leftrightarrow\dfrac{13}{4}=\dfrac{5x-7}{6}\)
\(\Leftrightarrow2\left(5x-7\right)=3.13\)
\(\Leftrightarrow10x-14=39\)
\(\Leftrightarrow10x=53\)
\(\Leftrightarrow x=5,3\)
\(c,\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)
\(\Leftrightarrow\dfrac{3x-9+11x+11}{33}=\dfrac{x+7-9}{9}\)
\(\Leftrightarrow\dfrac{14x+2}{33}=\dfrac{x-2}{9}\)
\(\Leftrightarrow33\left(x-2\right)=9\left(14x+2\right)\)
\(\Leftrightarrow33x-66=126x+18\)
\(\Leftrightarrow-93x=84\)
\(\Leftrightarrow x=-\dfrac{28}{31}\)
\(d,\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{3\left(3x-0,4\right)+2\left(1,5-2x\right)}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{9x-1,2+3-4x}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{5x+1,8}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow5\left(5x+1,8\right)=6\left(x+0,5\right)\)
\(\Leftrightarrow25x+9=6x+3\)
\(\Leftrightarrow19x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{19}\)
\(\Leftrightarrow77x=378\)
\(\Leftrightarrow x=\dfrac{54}{11}\)
ND
4 tháng 3 2018
a.
3x - 2 = 2x - 3
<=> 3x -2x = -3+2
<=> x = -1
Vậy.............
b.
\(5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
Vậy..........
30 tháng 4 2018
\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\)
\(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\)
\(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \)
Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\)
17 tháng 7 2022
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
HQ
15 tháng 4 2018
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
ND
7 tháng 7 2018
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
ND
7 tháng 7 2018
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
7 tháng 6 2017
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
7 tháng 6 2017
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
<=>\(\dfrac{\left(3x-0,4\right)15}{30}+\dfrac{\left(1,5-2x\right)10}{30}=\dfrac{\left(x+0,5\right)6}{30}\)
=>\(\left(3x-0,4\right)15+\left(1,5-2x\right)10=\left(x+0,5\right)6\)
<=>\(45x-6+15-20x=6x+3\)
<=>\(45x-20x-6x=6+3-15\)
<=>\(19x=-6\)
<=>\(x=\dfrac{-6}{19}\)