(3x-4)×(5x+15)=0
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18 tháng 3 2020
a) ( 5x - 4)(4x + 6)=0
<=> \([^{5x-4=0}_{4x+6=0}< =>[^{x=\frac{4}{5}}_{x=\frac{-6}{4}}\)
Vậy S = \(\left\{\frac{4}{5};\frac{-6}{4}\right\}\)
b) ( 3,5x - 7 )( 2,1x - 6,3 ) = 0
<=> \([^{3,5x-7=0}_{2,1x-6,3=0}< =>[^{x=2}_{x=3}\)
Vậy S = \(\left\{2;3\right\}\)
c) ( 4x - 10 )( 24 + 5x ) = 0
<=> \([^{4x-10=0}_{24+5x=0}< =>[^{x=\frac{5}{2}}_{x=\frac{-24}{5}}\)
Vậy S = \(\left\{\frac{5}{2};\frac{-24}{5}\right\}\)
d) ( x - 3 )( 2x + 1 ) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy S = \(\left\{3;\frac{-1}{2}\right\}\)
e) ( 5x - 10 )( 8 - 2x ) = 0
<=> \(\left[{}\begin{matrix}5x-10=0\\8-2x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy S = \(\left\{2;4\right\}\)
f) ( 9 - 3x )( 15 + 3x ) = 0
<=> \(\left[{}\begin{matrix}9-3x=0\\15+3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{3;-5\right\}\)
Học tốt nhaaa !
JS
15 tháng 2 2020
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
7 tháng 3 2022
\(a,3x-6=0\\ \Leftrightarrow x=2\\ b,\left(x+1\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\\ c,5x+3=2x+15\\ \Leftrightarrow5x-2x=15-3\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)
LM
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0
9 tháng 7 2018
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
TT
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
2
14 tháng 10 2021
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
14 tháng 10 2021
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
10 tháng 2 2020
1) Ta có: 3x-12=5x(x-4)
\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3x-12-5x^2+20x=0\)
\(\Leftrightarrow-5x^2+23x-12=0\)
\(\Leftrightarrow-5x^2+20x+3x-12=0\)
\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)
\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)
2) Ta có: 3x-15=2x(x-5)
\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)
3) Ta có: 3x(2x-3)+2(2x-3)=0
\(\Leftrightarrow\left(2x-3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)
4) Ta có: (4x-6)(3-3x)=0
\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)
10 tháng 2 2020
4) (4x - 6 ) ( 3 - 3x ) = 0
<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
\(\left(3x-4\right)\left(5x+15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-4=0\\5x+15=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4\\5x=-15\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-3\end{cases}}\)
Dấu \(\orbr{\begin{cases}\\\end{cases}}\)là dấu hoặc
Chúc bạn học tốt !!!
\(\left(3x-4\right).\left(5x+15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-4=0\\5x+15=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4\\5x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-3\end{cases}}}\)
Chúc bạn học tốt !