\(\frac{x}{4}=\frac{9}{12}\\ \Rightarrow x.12=4.9\\ \Rightarrow x=4.9:12\\ x=3\)
\(\frac{x}{4}=\frac{9}{12}\)
\(x=\frac{4.9}{12}\)
\(x=\frac{36}{12}\)
x=3
\(\frac{x}{4}=\frac{9}{12}=\frac{3}{4}\)
=> x = 3
Vậy x = 3
\(\frac{x}{4}=\frac{9}{12}\\ \Rightarrow x.12=4.9\\ \Rightarrow x=4.9:12\\ x=3\)
\(\frac{x}{4}=\frac{9}{12}\)
\(x=\frac{4.9}{12}\)
\(x=\frac{36}{12}\)
x=3
\(\frac{x}{4}=\frac{9}{12}=\frac{3}{4}\)
=> x = 3
Vậy x = 3
Tìm x biết
a) x+2x+3x+4x+...+100x=-213
b)\(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
c)3(x-2)+2(x-1)=10
d)\(\frac{x+1}{3}=\frac{x-2}{4}\)
e)\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
f)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
Tìm phân số \(\frac{x}{9}\)( x \(\in Z\)) sao cho
\(\frac{x}{9}< \frac{4}{7}< \frac{x+1}{9}\)
tìm x
\(\frac{x-1}{8}\)=\(\frac{5}{4}\)
\(\frac{6}{2x-1}\)=\(\frac{12}{-8}\)
\(\frac{2.x-5}{-12}\)=\(\frac{-6}{9}\)
Tìm x biết:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-(\frac{2}{3}-\frac{1}{4})\) \()\) Với x\(\in Z\)
Bài 2
a) Tìm x biết\(\frac{1}{2}-\left|\frac{5}{4}-2x\right|=\frac{1}{3}\)
b) Tìm x biết \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c) Tìm ba số x, y, z thỏa mãn: \(\frac{x}{y}=\frac{10}{9};\frac{y}{z}=\frac{3}{4}\)và \(x-y+z=78\)
Tìm x\(\in\)Q biết:
a,\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\)
b,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
Tìm x,biết: x2+\(\frac{2}{9}=\frac{5}{12}+\frac{1}{4}\)
1. Tìm x
\(\frac{\left(\frac{1}{9}\right)^x-\left(\frac{1}{3}\right)^x}{\left(\frac{1}{3}\right)^x}\)
2. Tìm các giá trị của x để các biểu thức sau không âm
\(\frac{x-1}{x^2+1}\)
3. Tính
\(\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
1.Tìm x, biết:(help me)
a)\(|x-1|—\left(-2\right)^3=9\times\left(-1\right)^{100}\)
b)\(\frac{x-2}{-4}\)=\(\frac{-9}{x-2}\)
c)\(\frac{x-5}{3}=\frac{-12}{5-x}\)
d)\(8x-|4x+\frac{3}{4}|=x+2\)
e) anh \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2019}{2020}\)