Tìm x : x nhân 9 -x nhân 2 = x = 56
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\(9x-5x+3x=56\)
<=> \(7x=56\)
<=> \(x=8\)
Vậy...
\(9x-x+3x=165\)
<=> \(11x=165\)
<=> \(x=15\)
Vậy....
\(7x+8x-x=168\)
<=> \(14x=168\)
<=> \(x=12\)
Vậy...
\(\frac{5}{8}x-\frac{1}{3}x-\frac{1}{6}x=15\)
<=> \(\frac{1}{8}x=15\)
<=> \(x=120\)
Vậy...
Dấu . là dấu nhân nha
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a, 9 . x - 5 . x + 3 . x = 56
x . (9 - 5 + 3) = 56
x . 7 = 56
x = 56 : 7
x = 8
b, 9 . x - x + 3 . x = 165
x . (9 - 1 + 3) = 165
x . 11 = 165
x = 165 : 11
x = 15
c, 7 . x + 8 . x - x = 168
x . (7 + 8 - 1) = 168
x . 14 = 168
x = 168 : 14
x = 12
d, 5/8 . x - 1/3 . x - 1/6 . x = 15
x . (5/8 - 1/3 - 1/6) = 15
x . 1/8 = 15
x = 15 : 1/8
x = 120
\(\left(-5\right)^2.x=56+22x\)
\(\Leftrightarrow25x=56+22x\)
\(\Leftrightarrow25x-22x=56\)
\(\Leftrightarrow3x=56\)
\(\Rightarrow x=\frac{56}{3}\)
Ta có :\(x+2x+3x+...+9x=459-9\)
\(\Rightarrow\)\(x\left(1+2+3+...+9\right)=450\)
\(\frac{\Rightarrow x.\left[\left(9-1\right):1+1\right].\left(9+1\right)}{2}\)\(=450\)
\(\frac{\Rightarrow x.90}{2}=450\)
\(\Rightarrow x.90=450.2=900\)
\(\Rightarrow x=\frac{900}{90}\)
\(\Rightarrow x=10\)
\(\text{Tìm }x\text{ : }\)
\(\text{Ta có : }\)
\(x+2x+3x+...+9x=459-9\)
\(\Rightarrow\text{ }x+2x+3x+...+9x=450\)
\(\Rightarrow\text{ }x\cdot\left(\left(9-1\right)\text{ : }1+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot\left(8\text{ : }1+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot\left(8+1\right)\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(\Rightarrow\text{ }x\cdot9\cdot\left(\left(9+1\right)\text{ : }2\right)=450\)
\(x\cdot9\cdot\left(10\text{ : }2\right)=450\)
\(x\cdot9\cdot5=450\)
\(x\cdot45=450\)
\(x=450\text{ : }45\)
\(x=10\)
`a. =4(x^2+4x+3)=4(x^2+3x+x+3)=4(x+3)(x+1)`
`b. =x^2+8x-7x-56=x(x+8)-7(x+8)=(x+8)(x-7)`
`c. =x^2-9x+8x-72=x(x-9)+8(x-9)=(x-9)(x+8)`
`d. =(x-y)^2-9=(x-y-3)(x-y+3)`
\(x^2.x=\left(-9\right).3\)
\(\Rightarrow x^3=-27\)
\(\Rightarrow x^3=\left(-3\right)^3\)
\(\Rightarrow x=-3\)
Vậy x = -3.
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)
=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)
=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)
=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)
=> \(3\left(8x-6\right)=9\left(x+4\right)\)
=> \(24x-18=9x+36\)
=> \(24x-18-9x=36\)
=> \(24x-9x=54\)
=> \(15x=54\)
=> \(5x=18\)
=> \(x=\frac{18}{5}\)
Vậy x = \(\frac{18}{5}\)
b) \(\left(x-\frac{1}{2}\right)\left(\frac{3}{2}-2x\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{2}-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)