2 x (1/4 - 3x) = 1/5 - 4x
ai biết giải hộ mk với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2x1/4-6x=1/5-4x
1/2-6x=1/5-4x
1/2-1/5=-4x+6x
3/10=2x
x=2:3/10
x=20/3
(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)
20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1
10x2−19x−33=010x2−19x−33=0
(10x+11)(x−3)=0
chỉ bt lm con b thoy
..army,,,,,,,,,,
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\frac{22}{5}\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2\)
\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)
\(\Leftrightarrow20x^2-16x-33=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
a) x-4=(4x+1)(x+2)
\(\Leftrightarrow x-4=4x^2+9x+2\)
\(\Leftrightarrow x-4-4x^2-9x-2=0\)
\(\Leftrightarrow-4x^2-8x-6=0\)
\(\Leftrightarrow-4\left(x^2+2x+\frac{3}{2}\right)=0\)
\(\Leftrightarrow x^2+2x+1+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x+1\right)^2=-\frac{1}{2}\)(vô lý)
Vậy: \(S=\varnothing\)
c) Ta có: |x-1|=x+5(*)
Trường hợp 1: \(x\ge1\)
(*)\(\Leftrightarrow x-1=x+5\)
\(\Leftrightarrow-6=0\)(vô lý)
Trường hợp 2: x<1
(*)\(\Leftrightarrow1-x=x+5\)
\(\Leftrightarrow1-x-x-5=0\)
\(\Leftrightarrow-2x-4=0\)
\(\Leftrightarrow-2x=4\)
hay x=-2(tm)
Vậy: S={-2}
d) Ta có: 3x-1=2
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
giải phương trình sau:
a, (3x+1/4)-1/3*(6x+9/5)=1
b, (5/2x+1)-(2x/1-2x)=1-(6-4x/4x^2-1)
giải hộ mk vs ạ
a, \(\left(3x+\frac{1}{4}\right)-\frac{1}{3}\left(6x+\frac{9}{5}\right)=1\)
\(3x+\frac{1}{4}-\frac{6}{3}x-\frac{3}{5}=1\)
\(x-\frac{7}{20}=1\Leftrightarrow x=\frac{27}{20}\)
b,ĐKXĐ : x \(\ne\)-1/2 ; 1/2
\(\left(\frac{5}{2x+1}\right)-\left(\frac{2x}{1-2x}\right)=1-\left(\frac{6-4x}{4x^2-1}\right)\)
\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{6-4x}{4x^2-1}\)
\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{2\left(3-2x\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(\frac{5\left(1-2x\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2x\left(2x+1\right)^2\left(2x-1\right)}{\left(1-2x\right)\left(2x+1\right)^2\left(2x-1\right)}=\frac{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2\left(3-2x\right)\left(2x+1\right)\left(1-2x\right)}{\left(2x+1\right)\left(2x-1\right)^2\left(2x-1\right)\left(1-2x\right)}\)
\(22x-5-20x^2-8x^3=18x-7-8x^3-4x^2\)
lm nốt nha,bị troll rồi ko vt đc nữa.
1. Ta có : 3x+12=0 <=> x= -4
bảng xét dấu:
x | -∞ -4 + ∞ |
3x+12 |
- 0 + |
f(x) >0 ∀ x ∈ (-4;+∞)
f(x) <0 ∀ x∈ (-∞;-4)
2. Ta có : -5x+9=0 <=> x= \(\frac{9}{5}\)
Bảng xét dấu:
x | -∞ 9/5 +∞ |
-5x+9 | + 0 - |
f(x) >0 ∀ x ∈ (-∞; 9/5)
f(x) <0 ∀ x ∈(9/5; +∞)
3. Ta có : -3x-9=0 <=> x= -3
x | -∞ -3 +∞ |
-3x-9 | + 0 - |
f(x) >0 ∀ x∈ (-∞; -3)
f(x) <0 ∀x∈ ( -3; +∞ )
4. Ta có : x (2x+4)=0
+, x=0
+, 2x+4=0 <=> x= -2
x | -∞ -2 0 +∞ |
x | - \(|\) - 0 + |
2x+4 | - 0 + \(|\) + |
f (x) | + 0 - 0 + |
f(x) >0 ∀ x ∈ (-∞; -2) \(\cup\) (0; +∞)
f(x) <0 ∀ x ∈ (-2;0)
5. Ta có: (x-2)(-x+4)=0
+, x-2=0 <=> x=2
+, -x+4=0 <=> x= 4
x | -∞ 2 4 +∞ |
x-2 | - 0 + \(|\) + |
-x+4 | + \(|\) + 0 - |
f(x) | - 0 + 0 - |
f(x) >0 ∀ x ∈ (2;4)
f (x) <0 ∀x∈ (-∞;2) \(\cup\)(4; +∞)
6. Ta có : (-4x+3)(x-6)=0
+, -4x+3=0 <=>x= \(\frac{3}{4}\)
+, x-6 =0 <=> x=6
x | -∞ 3/4 6 +∞ |
-4x+3 | + 0 - \(|\) - |
x-6 | - \(|\) - 0 + |
f(x) | - 0 + 0 - |
f(x) >0 ∀ x∈ (3/4;6)
f(x) <0 ∀ x∈ (-∞; 3/4) \(\cup\)(6;+∞)
a,(x+1)-(x+2)-(x+3)=24
=>x+1-x-2-x-3 =24
=>(x-x-x)+(1-2-3) =24
=> -x-4 =24
=> -x =24+4
=> -x =28
=> x =-28
Vậy x=-28
b,4x+2-3(x-1)=3x-5
=>4x+2-3x+3=3x-5
=>3x-4x+3x =2+3+5
=>2x =10
=>x =5
Vậy x=5
c,x-1-2(x-2)=x-11
=>x-1-2x+4=x-11
=>x-2x-x =-11+1-4
=>-2x =-14
=>x =7
Vậy x = 7
2 x (1/4 - 3x) = 1/5 - 4x
=> 1/2 . 6x = 1/5 - 4x
=> 1/2 - 1/5 = 6x - 4x
=> 3/10 = 2x
=> x = 3/20
tiick mk nha
2 \(\times\) (1/4 - 3x) = 1/5 - 4x
-> (2\(\times\) 1/4) - (2 \(\times\) 3x) = 1/5 - 4x
-> 1/2 - 6x = 1/5 - 4x
->1/2 - 6x - 1/5 + 4x = 0
->3/10 - 2x = 0
->2x = 3/10
->x = 3/20
Vậy....