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15 tháng 1 2017

2(x-3)-3(3-x)=15-3x
2x-6-9+3x=15-3x

2x+3x+3x=15+6+9
8x=30
x=30/8
x=3,75

16 tháng 1 2017

b) ( x + 3 )3 = -8

=> ( x + 3 )3 = (-2)2

=> x + 3 = -2

=> x = -2 - 3

=> x = -5

vậy x=-5

8 tháng 2 2020

a, Ta có : \(10+3\left(x-1\right)=10+6x\)

=> \(10+3x-3-10-6x=0\)

=> \(-3x-3=0\)

=> \(x=-1\)

b, Ta có : \(2\left(x-2\right)+3\left(3-x\right)=-4\)

=> \(2x-4+9-3x=-4\)

=> \(-x=-9\)

=> \(x=9\)

c, \(\left(\left|x-2\right|+1\right)\left(x-3\right)=0\)

TH1 : \(x-2\ge0\left(x\ge2\right)\)

=> \(\left|x-2\right|=x-2\)

Nên ta có phương trình : \(\left(x-2+1\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1\left(kTM\right)\\x=3\end{matrix}\right.\)

=> \(x=3\)

TH2 : \(x-2< 0\left(x< 2\right)\)

=> \(\left|x-2\right|=2-x\)

Nên ta có phương trình : \(\left(2-x+1\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}3-x=0\\x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=3\end{matrix}\right.\) ( ktm )

d, Ta có : \(\left(x+5\right)\left(-3x-15\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\-3x-15=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x=-5\end{matrix}\right.\)

8 tháng 2 2020

1) 10 + 3 . (x - 1) = 10 + 6x

10 + 3x - 3 = 10 + 6x

3x - 6x = 10 - 10 + 3

-3x = 0 + 3

-3x = 3

x = 3 : (-3)

x = -1

Vậy x = -1

2) 2 . (x - 2) + 3 . (3 - x) = -4

2x - 4 + 9 - 3x = -4

2x - 3x = -4 + 4 - 9

-x = 0 - 9

-x = -9

=> x = 9

Vậy x = 9

3) (Ix - 2I + 1) . (x - 3) = 0

=> (Ix - 2I + 1) = 0 hoặc (x - 3) = 0

Ix - 2I = 0-1 x = 0 + 3

Ix - 2I = -1( loại ) x = 3( thỏa mãn )

Vậy x = 3

4) (x + 5) . (-3x - 15) = 0

=> (x + 5) = 0 hoặc (-3x - 15) = 0

x = 0 - 5 -3x = 0 + 15

x = -5 -3x = 15

x = 15 : (-3)

x = -5

Vậy x = -5

Tick cho mk nha

26 tháng 3 2020

a) \(x-2=-6\)

\(x=-6+2\)

\(x=-4\)

b) \(15-\left(x-7\right)=-21\)

\(x-7=36\)

\(x=43\)

c) \(4.\left(3x-4\right)-2=18\)

\(4\left(3x-4\right)=20\)

\(3x-4=5\)

\(3x=9\)

\(x=3\)

d) \(\left(3x-6\right)+3=32\)

\(3x-6=29\)

\(3x=29+6\)

\(3x=35\)

\(x=\frac{35}{3}\)

e) \(\left(3x-6\right).3=32\)

\(3x-6=\frac{32}{3}\)

\(3x=\frac{32}{3}+6\)

\(3x=\frac{50}{3}\)

\(x=\frac{50}{9}\)

f) \(\left(3x-6\right):3=32\)

\(3x-6=96\)

\(3x=102\)

\(x=34\)

g) \(\left(3x-6\right)-3=32\)

\(3x-6=35\)

\(3x=41\)

\(x=\frac{41}{3}\)

h) \(\left(3x-2^4\right).7^3=2.7^4\)

\(\left(3x-2^4\right)=2.7=14\)

\(\left(3x-16\right)=14\)

\(3x=14+16=30\)

\(x=10\)

i) \(\left|x\right|=\left|-7\right|\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

k) \(\left|x+1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

l) \(\left|x-2\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

m) \(x+\left|-2\right|=0\)

\(x+2=0\)

\(x=-2\)

o) \(72-3\left|x+1\right|=9\)

\(3\left|x-1\right|=63\)

\(\left|x-1\right|=21\)

\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)

p) Ta có: \(\left|x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)

mà \(x+1< 0\)

\(\Rightarrow x-1=-3\)

\(\Rightarrow x=-2\)

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

hok tốt!!

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 11 2018

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)