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7/3 : 0,2x = 7/6 : 0,8
7/3 : 0,2x = 35/24
0,2x = 7/3 : 35/24
0,2x = 8/5
x = 8/5 : 0,2
x = 8
7/3:0.2x=7/6:0.8
7/3 : 0.2 x = 35/24
0.2 x = 35/24 : 7/3
0.2 x = 5/8
x=5/8 : 0.2
x= 25/8
Vậy x=25/8
=>\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}.\dfrac{5}{4}\)
=>\(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)
=>0,2\(.x=\dfrac{7}{3}:\dfrac{35}{24}\)
=>0,2\(.x=\dfrac{7}{3}.\dfrac{24}{35}\)
=>0,2.\(x=\dfrac{8}{5}\)
=>\(x=\dfrac{8}{5}:0,2\)
=>\(x=\dfrac{8}{5}.5\)
=>\(x=8\)
Vậy x=8
7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36
Nên theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6
\(\Rightarrow\)x=6.5=30
y=6.6=36
z=6.7=42
vậy x=30,y=36,z=42
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
\(a)\dfrac{x^2}{6}=\dfrac{36}{x}\)
\(=>x^3=36.6\)
\(=>x^3=6^3\)
\(=>x=6\)
(câu b thiếu dữ kiện)
áp dụng dãy tỉ số bằng nhau ta có
x/3=y/7=z/2=x+y+z/3+7+2=-16/12=-4/3
=>x/3=-4/3=>x=-4/3X3=-4
=>y/7=-4/3=>y=-4/3X7=-9,(3)
=>z/2=-4/3=>z=-4/3X2=-2(6)
ĐK: \(y\ne0\)
\(\dfrac{x+3}{6}=\dfrac{2x+1}{7}\Leftrightarrow7x+21=12x+6\\ \Leftrightarrow5x=15\Leftrightarrow x=3\\ \Leftrightarrow\dfrac{3+3}{6}=\dfrac{2+3\cdot3}{y}\Leftrightarrow\dfrac{11}{y}=1\Leftrightarrow y=11\)
Vậy \(x=3;y=11\)
1:
a: x/15=-2/6
=>x/15=-1/3
=>x=-5
b: 3/x=1,8/2
=>3/x=9/10
=>x=3*10/9=30/9=10/3
c: (x-3)/(x+2)=2/7
=>2x+4=7x-21
=>-5x=-25
=>x=5
d: (x+1)/3=(x-6)/8
=>8x+8=3x-18
=>5x=-26
=>x=-26/5
e: (2-x)/5=(x+4)/3
=>3(2-x)=5(x+4)
=>5x+20=6-3x
=>8x=-14
=>x=-7/4
g: (2x+1)/(-3)=(1-x)/2
=>2(2x+1)=3(x-1)
=>4x+2=3x-3
=>x=-5
a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}:0,8\)
⇔ \(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)
⇔ \(0,2x=\dfrac{7}{3}:\dfrac{35}{24}\)
⇔ \(0,2x=1,6\)
⇔ x = 1,6 : 0,2
⇔ x = 8
Vậy x =8