K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
3 tháng 3 2020
a) Ta có: \(\left(5x-15\right)\left(4+6x\right)=0\)
\(\Leftrightarrow5\left(x-3\right)\cdot2\cdot\left(2+3x\right)=0\)
\(\Leftrightarrow10\left(x-3\right)\left(2+3x\right)=0\)
Vì 10\(\ne\)0 nên
\(\left[{}\begin{matrix}x-3=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;\frac{-2}{3}\right\}\)
b) Ta có: \(\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\5x=6\\\frac{1}{2}x=\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=\frac{3}{4}:\frac{1}{2}=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{6}{5};\frac{3}{2}\right\}\)
c) Ta có: \(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\)
\(\Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=3\\x=\frac{25}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{4};\frac{25}{12}\right\}\)
d) Ta có: \(\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5\left(x-1\right)-\frac{3}{2}-\frac{\left(2-3\right)\left(x-1\right)}{3}\right]=0\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5x-5-\frac{3}{2}-\frac{-1\left(x-1\right)}{3}\right]=0\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(5x-5-\frac{3}{2}-\frac{1-x}{3}\right)=0\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(5x-\frac{13}{2}-\frac{1}{3}+\frac{x}{3}\right)=0\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{15x}{3}-\frac{41}{6}+\frac{x}{3}\right)=0\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{16x}{3}-\frac{41}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{1}{6}=0\\\frac{16x}{3}-\frac{41}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{1}{6}\\\frac{16}{3}\cdot x=\frac{41}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}:\frac{2}{3}\\x=\frac{41}{6}:\frac{16}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{41}{32}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{4};\frac{41}{32}\right\}\)
TN
3 tháng 3 2020
\(a.\left(5x-15\right)\left(4+6x\right)=0\\ \left[{}\begin{matrix}5x-15=0\\4+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)
\(b.\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}=0\right)\\ \left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=-\frac{3}{2}\end{matrix}\right.\)
c.
\(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\\ \Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\\ \left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{2}\end{matrix}\right.\)
3 tháng 5 2017
c. x^2-5x+6=0
<=> x^2-5x=-6
<=> -4x=-6
<=> x=-6/-4
vậy tập nghiệm của pt là s={-6/-4}
3 tháng 5 2017
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
DT
3 tháng 5 2017
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
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 !
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
<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
TN
10 tháng 4 2020
* 4x - 1 = 3x - 2
⇔ 4x - 3x = -2 + 1
⇔ x = -1
Vậy tập nghiệm của pt là S = {-1}
* \(\frac{3}{4}-3x=0\)
⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)
⇒ 3 - 12x = 0
⇔ 12x = 3
⇔ x = \(\frac{3}{12}=\frac{1}{4}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)
* 3x - 2 = 2x + 3
⇔ 3x - 2x = 3 + 2
⇔ x = 5
Vậy tập nghiệm của pt là S = {5}
* 2(x - 3) = 5(x + 4)
⇔ 2x - 6 = 5x + 20
⇔ 2x - 5x = 20 + 6
⇔ -3x = 26
⇔ x = \(\frac{-26}{3}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)
10 tháng 4 2020
\(A,5x-25=0\)
\(\Leftrightarrow5x-5^2=0\)
\(\Leftrightarrow5\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
Chúc bạn học tốt !
\(a,3x-6=0\\ \Leftrightarrow x=2\\ b,\left(x+1\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\\ c,5x+3=2x+15\\ \Leftrightarrow5x-2x=15-3\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)
thank you