Ôn tập toán 7
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Cho đề bài sau:\(\frac{1+x}{3}=\frac{8+2x}{3y}=\frac{3+x}{5}\). Hãy tìm giá trị x,y
Help me, please!! Thanks. Hope you do the exercise in decorum...:)
Tìm x biết :
a) \(-\frac{2}{3}.x+4=-12\)
b) \(-\frac{3}{4}+\frac{1}{4}:x=-3\)
c) \(\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
d)\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
Rút gọn A = \(\frac{x^2}{x^2-1}-\frac{x^2}{x^2+1}\left(\frac{x}{x+1}+\frac{1}{x^2+x}\right)\)
tìm x biết:
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
Tim x:
\(x-\frac{\frac{x}{2}-\frac{x}{3}}{4}-\frac{x}{6}=\frac{2\left(1+x\right)}{3}-\frac{\frac{x}{3}+\frac{1-x^7}{4}}{2}\)
Tìm x: \(x-\frac{\frac{x}{2}-\frac{x}{3}}{4}-\frac{x}{6}=\frac{2\left(1+x\right)}{3}-\frac{\frac{x}{3}+\frac{1-x^7}{4}}{2}\)
Tìm x: \(\frac{3\frac{1}{2}x-4}{6}+\frac{2+\frac{1}{4}x}{3}=1\frac{1}{4}x\left(x-1\right)-\frac{7-\frac{3}{4}x}{3}\)
Tìm GTNN của biểu thức : \(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\) biết \(x>0\).
Tìm x : \(\frac{\left(x-3\right)^2}{2}-1\frac{1}{3}\left(x+2\right)^2-\frac{5}{4}\left(x-1\right)\left(x+1\right)=1\frac{1}{2}x\left(x-2\right)-x-4\)