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4 tháng 2 2021
\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{5;-2\right\}\)
\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)
Câu d xem lại đề
H
20 tháng 5 2023
`5-(x-6)=4(3-2x)`
`<=>5-x+6-4(3-2x)=0`
`<=> 5-x+6-12 +8x=0`
`<=> 7x -1=0`
`<=> 7x=1`
`<=>x=1/7`
Vậy pt đã cho có nghiệm `x=1/7`
__
`3-x(1-3x) =5(1-2x)`
`<=> 3-x+3x^2=5-10x`
`<=> 3-x+3x^2-5+10x=0`
`<=> 3x^2 +9x-2=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{105}}{6}\\x=\dfrac{-9-\sqrt{105}}{6}\end{matrix}\right.\)
Vậy pt đã cho có tập nghiệm \(S=\left\{\dfrac{-9+\sqrt{105}}{6};\dfrac{-9-\sqrt{106}}{5}\right\}\)
__
`(x-3)(x+4) -2(3x-2)=(x-4)^2`
`<=>x^2+4x-3x-12- 6x +4 =x^2 -8x+16`
`<=>x^2-5x-8=x^2-8x+16`
`<=> x^2 -5x-8-x^2+8x-16=0`
`<=> 3x-24=0`
`<=>3x=24`
`<=>x=8`
Vậy pt đã cho có nghiệm `x=8`
20 tháng 5 2023
a) 5-(x-6)=4(3-2x)
=> 5 – x + 6 = 12 – 8x
=> -x + 8x = 12 – 5 – 6
=> 7x = 1
=> x=1/7
Vậy phương trình có nghiệm x=1/7
b) 3 - x ( 1 - 3x)=5(1-2x)
=> 3-x+3x^2=5-10x
=> 3x^2+9x-2= 0
0=105
=> x =\(\dfrac{-9-\sqrt{105}}{6}\)
TM
11 tháng 8 2021
1/ \(2\left(x-5\right)=\left(-x-5\right)\)
\(\Leftrightarrow2x-10=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
==========
2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy: \(S=\left\{7\right\}\)
==========
3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1=x-19\)
\(\Leftrightarrow0x=0\)
Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\)
===========
4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2=10-15x\)
\(\Leftrightarrow14x=1\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)
==========
5/ \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x=7x+1\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy: \(S=\left\{-3\right\}\)
[---]
Chúc bạn học tốt.
NT
11 tháng 8 2021
1. \(2\left(x-5\right)=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy \(S=\left\{\dfrac{5}{3}\right\}\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1-x+19=0\)
\(\Leftrightarrow0x=0\)
Vậy \(S=\left\{x\in R\right\}\)
4. \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2-10+15x=0\)
\(\Leftrightarrow14x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy \(S=\left\{\dfrac{1}{14}\right\}\)
4. \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x-7x-1=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
PT
1
PT
1
CM
21 tháng 11 2019
Ta có: ( x - 2 )( 3x + 5 ) = ( 2x - 4 )( x + 1 )
⇔ ( x - 2 )( 3x + 5 ) - 2( x - 2 )( x + 1 ) = 0
⇔ ( x - 2 )[ ( 3x + 5 ) - 2( x + 1 ) ] = 0
⇔ ( x - 2 )( x + 3 ) = 0
Vậy phương trình đã cho có tập nghiệm là S = { - 3;2 }.
Y
12 tháng 4 2022
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
Y
12 tháng 4 2022
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
TM
11 tháng 8 2021
1/ \(7x-5=13-5x\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)
==========
2/ \(19+3x=5-18x\)
\(\Leftrightarrow21x=-14\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)
==========
3/ \(x^2+2x-4=-12+3x+x^2\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
===========
4/ \(-\left(x+5\right)=3\left(x-5\right)\)
\(\Leftrightarrow-x-5=3x-15\)
\(\Leftrightarrow-4x=-10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)
==========
5/ \(3\left(x+4\right)=\left(-x+4\right)\)
\(\Leftrightarrow3x+12=-x+4\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy: \(S=\left\{-2\right\}\)
[----------]
NT
11 tháng 8 2021
1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)
2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)
3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)
4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)
5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)
PN
0
9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
1) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
2) \(9x^2-1=3x+1\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{2}{3}\end{cases}}\)