1)(2x-5)(x+9)+6x=30
2)(x+3)(5x-15)+2x-6=0
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D
3 tháng 11 2018
\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)
NT
1
10 tháng 11 2018
\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
JS
15 tháng 2 2020
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
15 tháng 3 2020
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
15 tháng 3 2020
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
28 tháng 9 2018
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
TL
1
12 tháng 5 2023
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
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.
\(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\\ \Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\\ \left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{2}\end{matrix}\right.\)
15 tháng 2 2020
Mấy cái này chuyển vế đổi dấu là xong í mà :3
1,
16-8x=0
=>16=8x
=>x=16/8=2
2,
7x+14=0
=>7x=-14
=>x=-2
3,
5-2x=0
=>5=2x
=>x=5/2
Mk làm 3 cau làm mẫu thôi
Lúc đăng đừng đăng như v :>
chi ra khỏi ngt nản
10 tháng 3 2020
từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại
2) \(\left(x+3\right)\left(5x-15\right)+2x-6=0\)
\(\Leftrightarrow\left(x+3\right).5.\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[5\left(x+3\right)+2\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+17\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)
câu 1 phải là \(\left(x-5\right)\left(x+9\right)+6x=30\)
\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6x-30=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+9+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-15\end{matrix}\right.\)
tìm hả bn