![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 17 - 14(x + 1) = 13 - 4(x + 1) - 5(x - 3)
<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15
<=> -14x + 3 = -9x + 24
<=> -14x + 9x = 24 - 3
<=> -5x = 21
=> x = -4,2
Ta có : 5x + 3,5 + (3x - 4) = 7x - 3(x - 0,5)
<=> 5x + 3,5 + 3x - 4 = 7x - 3x + 1,5
<=> 8x - 0,5 = 4x + 1,5
=> 8x - 4x = 1,5 + 0,5
=> 4x = 2
=> x = \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4(3x – 2) – 3(x - 4) = 7x + 20.
⇔ 12x – 8 – 3x + 12 = 7x + 20
⇔ 9x – 7x = 20 + 8 – 12
⇔ 2x = 16
⇔ x = 8
Vậy phương trình có tập nghiệm S = { 8}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)
\(10x-15-20x+28=19-2x-22\)
\(10x-20x+2x=19-22-28+15\)
\(-8x=-16\)
\(\Rightarrow x=2\)
\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)
\(4x+12-7x+17=40x-8+166\)
\(4x-7x-40x=-8+166-17-12\)
\(-43x=129\)
\(x=-3\)
\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)
\(17-14x+14=13-4x-4-5x+15\)
\(-14x+4x+5x=13-4+15-14-17\)
\(-5x=-7\)
\(x=\frac{7}{5}\)
\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)
\(5x+3,5+3x-4=7x-3x+1,5\)
\(5x+3x-7x+3x=1,5-3,5\)
\(x=-2\)
\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)
\(28x+21-4x+4=15x+11,25+7\)
\(28x-4x-15x=11,25+7-4-21\)
\(9x=\frac{-27}{4}\)
\(x=\frac{-3}{4}\)
\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)
\(3x+0,8x+0,2x=3,38-2,42\)
\(4x=\frac{24}{25}\)
\(x=\frac{6}{25}\)
chúc bạn học tốt !!
![](https://rs.olm.vn/images/avt/0.png?1311)
5 x - 1 + 2 6 - 7 x - 1 4 = 2 2 x + 1 7 - 5 ⇔ 5 x - 3 6 - 7 x - 1 4 = 4 x + 2 7 - 5
⇔ 14(5x – 3) – 21(7x – 1) = 12(4x + 2) – 5.84
⇔ 70x – 42 – 147x + 21 = 48x + 24 – 420
⇔ 70x – 147x – 48x = 24 – 420 + 42 – 21
⇔ -125x = -375
⇔ x = 3
Phương trình có nghiệm x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(S=\left\{-2\right\}\)
[----------]
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. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Chuyển vế và rút gọn được 4x = 16, tìm được x = 4.
b) Đưa PT về dạng 5x = 15, tìm được x = 3
c) Quy đồng, khử mẫu thu được 6x - 9 + 24 = 2 - 2x.
Từ đó tìm được x = - 13 8
d) Quy đồng, khử mẫu thu được 30x + 9 = 36 + 24x +32
Từ đó tìm được x = 59 6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\dfrac{5x-5+2}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}-5\)
\(\Leftrightarrow\dfrac{70x-70+28}{84}-\dfrac{147x-21}{84}=\dfrac{48x+24}{84}-\dfrac{420}{84}\)
\(\Leftrightarrow70x-70+28-147x+21=48x+24-420\)
\(\Leftrightarrow-125x=-375\)
\(\Leftrightarrow x=3\)
g) \(\dfrac{5\left(x+1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
<=>\(\dfrac{14\left(5x-3\right)-21\left(7x-1\right)}{84}=\dfrac{12\left(4x-33\right)}{84}\)
<=>\(70x-42-147x+21=48x-396\)
<=>\(125x=375\)
<=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
4(x – 4) = -7x +17
⇔ 4x - 16 = -7x + 17
⇔ 4x + 7x = 17 + 16
⇔ 11x = 33
⇔ x = 3
Phương trình có tập nghiệm S = { 3}