3x + 3x+1 + 3x+2 = 117
Tìm x biết 3x+3x+1+3x+2=117
\(3x+3x+1+3x+2=117\)
\(\Rightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Rightarrow9x+3=117\)
\(\Rightarrow9x=117-3\)
\(\Rightarrow9x=114\)
\(\Rightarrow x=114:9\)
\(\Rightarrow x=\frac{38}{3}\)
Vậy \(x=\frac{38}{3}\)
P/s : Đúng nha
~ Ủng hộ nhé
a) 3x + 3x+1 + 3x+2 =117
b) 3 + 4 (x - 10) = 32 + 6
a)
\(3^x+3^{x+1}+3^{x+2}=117\\ \Leftrightarrow3^x+3.3^x+9.3^x=117\\ 13.3^x=117\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)
b)
\(3+4\left(x-10\right)=3^2+6\\ \Leftrightarrow3+4\left(x-10\right)=15\\ \Leftrightarrow4\left(x-10\right)=12\\ \Leftrightarrow x-10=3\\ \Leftrightarrow x=13\)
a) \(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x.\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(x=2\)
b) \(3+4\left(x-10\right)=3^2+6\)
\(3+4x-40=9+6\)
\(4x=15+40-3\)
\(4x=52\)
\(x=13\)
\(3x+3x+1+3x+2=117\)
tìm x
3x+3x+1+3x+2=177
9x+3=117
9x=117-3
9x=114
x=114/9
\(3x+3x+1+3x+2=117\)
\(\Leftrightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Leftrightarrow9x+3=117\)\(\Rightarrow9x=114\Rightarrow x=\frac{114}{9}\)
\(\text{Vậy x=}\frac{114}{9}\)
3x+3x+1+3x+2=117
=>3x.3 +(1+2) = 117
=>9x +3 = 117
=> 9x = 114
=> x = \(\frac{38}{3}\)
Tìm \(x:\)
\(3x+3x+1+3x+2=117\)
\(3x+3x+1+3x+2=117\)
\(\Rightarrow3x+3x+3x=117-1-2\)
\(\Rightarrow3x+3x+3x=114\)
\(\Rightarrow x.\left(3+3+3\right)=114\)
\(\Rightarrow x.9=114\)
\(\Rightarrow x=\dfrac{38}{3}\)
Vậy \(x=\dfrac{38}{3}\)
=> 3x+3x+3x+1+2=117
=>9x+3=117
=>9x=117-3=114
=> x=\(\dfrac{114}{9}\)
\(3x+3x+1+3x+2=117\)
\(\Leftrightarrow9x=114\)
\(\Leftrightarrow x=\dfrac{38}{3}\)
bài 1: Tính giá trị biểu thức
A = x(3x-y)-(3x+1)y tại x = 4/3; y = -1
B = \(3\frac{1}{117}.\frac{1}{119}-\frac{4}{117}.5\frac{118}{119}-\frac{8}{39}\)
Bài 2: Tìm m và n để hai đa thức đồng nhất:
f(x)=(m-1)x^2+3x+1
g(x) = x^2-nx+1
Bài 1:
Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:
\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)
\(A=\frac{20}{3}+5=\frac{35}{3}\)
Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)
\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)
\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)
Tính giá trị biểu thức
D=3x^2(5x^2-4)+x^2(8-15x^2)-8x với |x|=3
F=(3+1/117).1/119-4/117. (5+118/119) - 5/(117.119) + 8/39
Tính giá trị biểu thức
D=3x^2(5x^2-4)+x^2(8-15x^2)-8x với |x|=3
F=(3+1/117). 1/119-4/117.(5+118/119)-5/119+8/39
tìm x y biết x/5 = x/6 và -3x - 4y = -117
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{-3x}{-15}=\dfrac{4y}{24}\)
Áp dụng t/c của DS bằng nhau, ta có: \(\dfrac{-3x-4y}{-15-24}=\dfrac{-117}{-39}=3\)
\(\dfrac{-3x}{-15}=3\Rightarrow x=15\)
\(\dfrac{4y}{24}=3\Rightarrow y=18\)
3x=4y=6z và x+y+z=117
\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{3}=\frac{z}{2}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{2}=\frac{x+y+z}{9}=\frac{117}{9}=13\)
\(\Rightarrow x=4.13=52;y=3.13=39;z=2.13=26\)