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1 tháng 12 2021

dấu [] là giá trị tuyệt đối nha

 

29 tháng 6 2021

a) 2x . 4 = 128

<=> 2x = 32 

<=> 2x = 25

<=> x = 5

b) x15 = x1

<=> x15 - x = 0

<=> x(x14 - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)3 = 125

<=> (2x + 1)3 = 53

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)4 = (x - 5)6

<=> (x - 5)6 - (x - 5)4 = 0

<=> (x - 5)4[(x - 5)2 - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5

Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

29 tháng 6 2021

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)

  Vậy ...

2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)

  Vậy ...

3) PT \(\Rightarrow3x-4-2x+5=3\)

          \(\Rightarrow x=2\)

 Vậy ...

4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

  Vậy ...

3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)

\(\Leftrightarrow3x-4-2x+5=3\)

\(\Leftrightarrow x+1=3\)

hay x=2

3 tháng 2 2021

1/ 3-2x+4+6x=x+7+3x

⇔-2x+6x-x-3x=0

⇔0x=0 (Vô số nghiệm)

2/-6(1,5-2x)=3(-15+2x)

⇔-9+12x=-45+6x

⇔6x+36=0

⇔6(x+6)=0

⇔x+6=0

⇔x=-6

Vậy S ϵ {-6}

3/ 3(2x-5)+5(x-1)=4(x+1)

⇔6x-15+5x-5=4x+4

⇔7x=24

⇔x=\(\dfrac{24}{7}\) 

Vậy S ϵ {\(\dfrac{24}{7}\)}

 

1) Ta có: \(3-2x+4+6x=x+7+3x\)

\(\Leftrightarrow4x+7=4x+7\)

\(\Leftrightarrow4x+7-4x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

2) Ta có: \(-6\cdot\left(1.5-2x\right)=3\left(-15+2x\right)\)

\(\Leftrightarrow-9+12x=-45+6x\)

\(\Leftrightarrow12x-9+45-6x=0\)

\(\Leftrightarrow6x+36=0\)

\(\Leftrightarrow6x=-36\)

hay x=-6

Vậy: S={-6}

3) Ta có: \(3\left(2x-5\right)+5\left(x-1\right)=4\left(x+1\right)\)

\(\Leftrightarrow6x-15+5x-5=4x+4\)

\(\Leftrightarrow11x-20-4x-4=0\)

\(\Leftrightarrow7x-24=0\)

\(\Leftrightarrow7x=24\)

\(\Leftrightarrow x=\dfrac{24}{7}\)

Vậy: \(S=\left\{\dfrac{24}{7}\right\}\)

31 tháng 8 2021

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

31 tháng 8 2021

Sorry, cái này mình nhầm

 

8 tháng 7 2016

- Mình có thể giúp bài 1 :)

8 tháng 7 2016

 5^x +  5^ ( x + 2 ) = 650

5+  5x . 52 = 650

 5.( 1 + 25 ) = 650

 5x . 26 = 650 

5x = 650 : 26

 5x = 25

 5x = 52 

=> x = 2 

Vậy x = 2 

8 tháng 1 2018

x(5 – 2x) + 2x(x – 1) = 15

(x.5 – x.2x) + (2x.x – 2x.1) = 15

5x – 2x2 + 2x2 – 2x = 15

(2x2 – 2x2) + (5x – 2x) = 15

3x = 15

x = 5.

Vậy x = 5.

30 tháng 11 2016

\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

\(x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(x-2\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)

\(\left(2x-3\right)\left(x^2+4\right)=0\)

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

\(x\left(x-1\right)+5x-5=0\)

\(x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(x+5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)

\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

30 tháng 11 2016

Cảm ơn

 

19 tháng 7 2016

x . (5 - 2x) + 2x . (x - 1) = 15

5x - 2x2 + 2x2 - 2x = 15

(5x - 2x) + (2x2 - 2x2) = 15

3x = 15

x = 15 : 3

x = 5

 
19 tháng 7 2016

 

x.(5-2x)+2x.(x-1)=15

<=>5x-2x2+2x2-2x=15

<=>3x=15

<=>x=5

 

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

 

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)