1)(2x-5)(x+9)+6x=30
2)(x+3)(5x-15)+2x-6=0
Những câu hỏi liên quan
1,x=3x2
2,(x+5)(x-3)-(x-30)=0
3,(2x-6)(x+4)+2(2x-6)=0
4,(2x-5)(x+9)+6x-15=0
3,(2x-5)(x+6)+8x-20=0
\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)
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1)2x-8x2=0
2)x-x2=0
3)7x=5x2
4)3x2=x
5) (2x-5)(x+9)+8-6x=0
6) (x+3)(5x-15)+2x-6=0
\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
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1) (x+6)(3x-1)+x+6=0
2) (x+4)(5x+9)-x-4=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
4)2x (2x-3)=(3-2x)(2-5x)
5)(2x-7)^2-6(2x-7)(x-3)=0
6)(x-2)(x+1)=x^2-4
7) x^2-5x+6=0
8)2x^3+6x^2=x^2+3x
9)(2x+5)^2=(x+2)^2
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)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
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Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
https://i.imgur.com/PCDykdb.jpg
1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2
Vậy phương trình có nghiệm là x = 2
Xem thêm câu trả lời
1) (4-3x) (10x-5)02) (7-2x) (4+8x) 03) (9-7x) (11-3x) 04) (7-14x) (x-2) 05) (2x+1) (x-3) 06) (8-3x) (-3x+5) 07) (16-8x) (2-6x) 08) (x+4) (6x-12) 09) (11-33x) (x+11) 010) (x-1/4) (x+5/6) 011) (7/8-2x) (3x+1/3) 012) 3x - 2x^2 013) 5x + 10x^2 014) 4x + 3x^2 015) -8x^2 + x 016) 10x^2 - 15x 017) x^2 -4 018) 9 - x^2 019) x^2 -1 020) (x-3) (2x-1) (2x-1) ( 2x+3)21) (5+4x) (-x+2) (5+4x) (7+5x)22) (4+x) (x-5) (3x-8) (x-5) 0 23) (3x-8) (7-21x) - (9+2x) (7-21x)24) (10+ 7x) (x+1) (9x-2)(...
Đọc tiếp
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
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
Tìm x
1. (x-2)2 - (x+3)2 - 4(x+1) = 5
2. (2x-3)(2x+3) - 3x(x-5) = -44
3. (5x+1)2 - (5x+3)(5x-3) = 30
4. (x+3)(x2-3x+9) - x(x-2)(x+2) = 15
5. (x+2)3 - x(x-3)(x+3) - 6x2 = 29
6. x3 - 16x =0
7. (x-1)(x+2)-x-2 = 0
8. x(2x-3) - 2(3-2x) = 0
9. 2x(x-3) - x+3 = 0
10. (2x-1)2 — (1-2x)= 0
Giải giúp mình bài này nha
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
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Giải pT sau : a.x(4x-1)^2(2x-1)=9 b.(x^2+5x+6)(x^2-11x+30)=180 c.6x^4-5x^3-38x^2-5x+6=0
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
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
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a) (5x-15)(4+6x)=0
b) (2x-1)(5x-6)(1/2x-3/4)=0
c) (3-4x)(2x-3/4-x-4/3)=0
d) (2/3x-1/6)[5(x-1)-3/2-(2-3)(x-1)/3]=0
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: \(x\in\left\{\frac{1}{4};\frac{41}{32}\right\}\)
\(a.\left(5x-15\right)\left(4+6x\right)=0\\ \left[{}\begin{matrix}5x-15=0\\4+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)
\(b.\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}=0\right)\\ \left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=-\frac{3}{2}\end{matrix}\right.\)
c.
\(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\\ \Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\\ \left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{2}\end{matrix}\right.\)
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
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
từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại
1, 16 - 8x = 0
<=>-8x = 16
<=> x = -2
Vậy_
2, 7x + 14 = 0
<=> 7x = -14
<=> x = -2
3, 5 - 2x = 0
<=> - 2x = -5
<=> x =\(\frac{5}{2}\)
Vậy_
4, 3x - 5 = 7
<=> 3x = 7 + 5
<=> 3x = 12
<=> x = 4
Vậy...
5, 8 - 3x = 6
<=> - 3x = 6 - 8
<=> -3x = - 2
<=> x =\(\frac{2}{3}\)
Vậy......