a) \(128-3\left(x+4\right)=23\)

\(\Rightarrow3\left(x+4\right)=128-23\)

\(\Rightarrow3\left(x+4\right)=105\)

\(\Rightarrow x+4=35\)

\(\Rightarrow x=35-4\)

\(\Rightarrow x=31\)

b) \(\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\Rightarrow\left(4x+28\right)\cdot3+55=35\cdot5\)

\(\Rightarrow\left(4x+28\right)\cdot3+55=175\)

\(\Rightarrow\left(4x+28\right)\cdot3=120\)

\(\Rightarrow4x+28=40\)

\(\Rightarrow4x=12\)

\(\Rightarrow x=3\)

a, \(128-3\left(x+4\right)=23\)

\(=>3\left(x+4\right)=128-23\)

\(=>3\left(x+4\right)=105\)

\(=>x+4=105:3\)

\(=>x+4=35\)

\(=>x=35-4\)

\(=>x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(=>\left(4x+28\right).3+55=35.5\)

\(=>\left(4x+28\right).3+55=175\)

\(=>\left(4x+28\right).3=175-55\)

\(=>\left(4x+28\right).3=120\)

\(=>4x+28=120:3\)

\(=>4x+28=40\)

\(=>4x=40-28\)

\(=>4x=12\)

\(=>x=12:4\)

\(=>x=3\)

\(#WendyDang\)

a) 128 - 3 (x + 4) =23

      3.( x + 4) = 128 -23

      3. ( x + 4) =  105

      x + 4 = 105 :3

      x + 4 =  35

      x = 35 -4

     x = 31

a) `4x(x-5)-(x-1)(x-3)=23`

`<=> 4x^2-20x-x^2+4x-3=23`

`<=>3x^2-16x-26=0`

`<=> x=(8+-\sqrt142)/3

*Nếu đề là: `4x(x-5)-(x-1)(4x-3)=23`

`<=> 4x^2-20x-4x^2+7x-3=23`

`<=>-13x=26`

`<=>x=-2`

b) `(x-5)(x-4)-(x+1)(x-2)=7`

`<=>x^2-9x+20-x^2+x+2=7`

`<=>-8x=-15`

`<=>x=15/8`

\(\text{a) 4x(x-5)-(x-1)(1x-3)=23}\)

\(\Leftrightarrow4x^2-20x-\left(x^2-3x-x+3\right)=23\)

\(\Leftrightarrow4x^2-20x-x^2+3x+x-3=23\)

\(\Leftrightarrow3x^2-16x-26=0\)

Giải phương trình bậc 2 một ẩn ta được :

\(x_1=\dfrac{8-\sqrt{142}}{3};x_2=\dfrac{8+\sqrt{142}}{3}\)

1)

\(x-17=23\\ \Rightarrow x=23+17\\ \Rightarrow x=40\)

2)

\(2\left(x-1\right)=7+\left(-3\right)\\ \Rightarrow2x-2=4\\ \Rightarrow2x=4+2\\ \Rightarrow2x=8\\ \Rightarrow x=4\)

3)

\(4\left(x+5\right)^3-7=101\\ \Rightarrow4\left(x+5\right)^3=101+7\\ \Rightarrow4\left(x+5\right)^3=108\\ \Rightarrow\left(x+5\right)^3=108\div4\\ \Rightarrow\left(x+5\right)^3=27\\ \Rightarrow\left(x+5\right)^3=3^3\\ \Rightarrow x+5=3\Rightarrow x=3-5\\ \Rightarrow x=-2\)

4)

\(2^{x+1}\times3+15=39\\ \Rightarrow2^{x+1}\times3=39-15\\ \Rightarrow2^{x+1}\times3=24\\ \Rightarrow2^{x+1}=24\div3\\ \Rightarrow2^{x+1}=8\)

\( \Rightarrow2^{2+1}=8\\\Rightarrow2^3=8\Rightarrow x=2 \)

a) x - 17 = 23

x = 23 + 17

x = 40

Vậy x = 40

b) 2 ( x - 1 ) = 7 + ( - 3 )

x - 1 = 4 : 2

x = 2 + 1

x = 3

Vậy x = 3

c) 4 ( x + 3 )^3 - 7 = 101

4 ( x + 3 )^3 = ( 101 + 7 ) : 4

( x + 3 )^3 = 3^3

⇒ x + 3 = 3

⇒ x = 0

Vậy x = 0

d) 2^{ x+ 1 } . 3 + 15 = 39

2^{ x + 1 } = ( 39 - 15 ) : 3

2^{ x + 1 } = 2{ 2 + 1 }

⇒ x + 1 = 2 + 1

⇒ x = 2

Vậy x = 2

a)x+5=20-5

x+5=15

x=15-5

x=10

b) sai đề

a)x+5=20-(12-7)

x+5=20-5

x+5=15

x=15-5

x=10

b)4*(3*x-4)=18

3*x-4=18:4

3*x-4=4,5

3*x=4,5+4

3*x=8,5

x=8,5:3

x=17/6

c)128-3*x(x+4)=18

3*x(x+4)=128-18

3*x(x+4)=110

4*x+12=110

4*x=110-12

4*x=98

x=98:4

x=24,5

d)60-3*(x-2)=51

3*(x-2)=60-51

3*(x-2)=9

x-2=9:3

x-2=3

x=3+2

x=5

a.

\(1,25\times4-x=2,5\)

\(5-x=2,5\)

\(x=5-2,5\)

\(x=2,5\)

b.

\(\left(158-x\right)\div7=20\)

\(158-x=20\times7\)

\(158-x=140\)

\(x=158-140\)

\(x=18\)

c.

\(x-0,25=\frac{43}{4}-\frac{18}{5}\)

\(x-\frac{1}{4}=\frac{43}{4}-\frac{18}{5}\)

\(x=\frac{43}{4}-\frac{18}{5}+\frac{1}{4}\)

\(x=\left(\frac{43}{4}+\frac{1}{4}\right)-\frac{18}{5}\)

\(x=\frac{44}{4}-\frac{18}{5}\)

\(x=11-\frac{18}{5}\)

\(x=\frac{37}{5}\)

a. 1,25 × 4 - x = 2,5

5- x= 2,5

x= 5-2,5

x=2,5

b.(158- x) :7= 20

158- x= 20x7

158 -x= 140

x=158- 140

x=18

c. x- 0,25= \(\frac{43}{4}+\frac{18}{5}\)

x- 0,25=14,35

x= 14,35+ 0,25

x=14,6

\(\Rightarrow\) 510984 = 3384 x (158 - x)

\(\Rightarrow\) 158 - x = 510984:3384

\(\Rightarrow\) 158 - x = 151

\(\Rightarrow\) x = 158 - 151 = 7

846 x 151 x 4 = 3384 x ( 158 - x )

<=> x = 7

7 nha ban