n: (-8)*x+17=-23

=>\(x\cdot\left(-8\right)=-23-17=-40\)

=>\(x\cdot8=40\)

=>\(x=\dfrac{40}{8}=5\)

o: \(\left(-15\right)\cdot x=10\left(-4\right)-5\)

=>\(x\cdot\left(-15\right)=-40-5=-45\)

=>\(x\cdot15=45\)

=>\(x=\dfrac{45}{15}=3\)

p: \(\left(-3\right)\cdot x-4=2\cdot\left(-7\right)+4\)

=>\(\left(-3\right)\cdot x-4=-14+4=-10\)

=>\(x\left(-3\right)=-10+4=-6\)

=>\(3x=6\)

=>\(x=\dfrac{6}{3}=2\)

q: x+x+x+91=-2

=>\(3x+91=-2\)

=>\(3x=-2-91=-93\)

=>\(x=-\dfrac{93}{3}=-31\)

r: \(-152-\left(3x+1\right)=\left(-2\right)\cdot\left(-27\right)\)

=>\(-152-3x-1=54\)

=>\(-153-3x=54\)

=>\(3x=-153-54=-207\)

=>\(x=-\dfrac{207}{3}=-69\)

Bài 1

a) -(515-80+91)-(2003+80-91)= -515+80-91-2003-80+91=(80-80)+(-91+91)+(-515-2003)=0+0+(-2518)= -2518

b)-(-85)-(-71)+15+(-85)=85+71+15+-85=(85-85)+(71+15)=0+86=86

c)-(537-812)+(-163-712)-(-364)= -537+812+-163-712+364=(812-712)+(-537-136)+364=0+(-700)+364= -236

Bài 2

a)25-(x+4)= -19+3                     c) I2x-5l=13

  25-(x+4)= -16                            suy ra 2x-5 thuộc {13;-13}

      (x+4)= 25-(-16)                     TH1: 2x-5=13 vậy x=9

       x+4 =41                             TH2: 2x-5= -13 vậy x= -4

       x     =41-4

       x     =37   

bài 2: c) x= 9

             x= -9

tick nha

mhinf là muốn loạn luôn

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Bài 1:

a) \(24 - (-15) - 2\)

\(=39-2\)

\(=37\)

b) \((-85) + 10 - (-85) - 50\)

\(=[(-85)-(-85)]+10-50\)

\(=0+10-50\)

\(=10-50\)

\(=-40\)

c) \(71 - (-30) - (+18) + (-30)\)

\(=[(-30)-(-30)]+71-(+18)\)

\(=0+71-18\)

\(=71-18\)

\(=53\)

d) \(-(30) - (+37) + (+37) + (-85)\)

\(=[-(+37)+(+37)]-(30)+(-85)\)

\(=0-(30)+(-85)\)

\(=(-30)+(-85)\)

\(=-115\)

e) \((35-815) - (795-65)\)

\(=(-780)-730\)

\(=-1510\)

g) \((2002-79+15) + (-79+15)\)

\(=1938+(-64)\)

\(=1874\)

Bài 2:

a) \(25 - (30+x) = x - (27-8)\)

\(25-30-x=x-27+8\)

\(x+x=25-30+27-8\)

\(2x=14\)

\(x=14\div2\)

\(x=7\)

b) \((x-12) - 15 = (20-17) - (18+x) \)

\(x-12-15=13-18-x\)

\(x-27=-5-x\)

\(x+x=-5+27\)

\(2x=22\)

\(x=22\div2\)

\(x=11\)

c) \(15 - x = 7 - (-2)\)

\(15-x=9\)

\(x=15-9\)

\(x=6\)

d) \(x - 35 = (-12) - 3\)

\(x-35=-15\)

\(x=-15+35\)

\(x=20\)

e) \(\left|5-x\right|-26=-15\)

\(\left|5-x\right|=-15+26\)

\(\left|5-x\right|=11\)

Từ đây ta có:

*Nếu \(5-x=11\)

\(x=5-11\)

\(x=-6\)

*Nếu \(5-x=-11\)

\(x=5-(-11)\)

\(x=16\)

Vậy \(x=-6;x=16\)

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

a) Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{25}{5}=5\) ( do x + y = 25 )

\(\Rightarrow\hept{\begin{cases}x=5.2=10\\y=5.3=15\end{cases}}\)

b) Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{91}=\frac{y}{15}=\frac{z}{6}=\frac{-x+y-z}{-91+15-6}=\frac{60}{-82}=\frac{-30}{41}\)( do -x + y - z = 60 )

Từ đó tìm được x , y , z

`16/803+38+(-16/803)`
`=16/803-16/803+38`
`=0+38=38`

`100/91+310-9/91`
`=100/91-9/91+310`
`=1+310=311`

\(\dfrac{16}{103}+\left(38+\dfrac{-16}{103}\right)\)

\(=\dfrac{16}{103}+38+\dfrac{-16}{103}\)

\(=\dfrac{16}{103}+\dfrac{-16}{103}+38\)

\(=0+38\)

\(=38\)