\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=0\)

\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)

\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

\(\Rightarrow x-100=0\left(Vì\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\right)\)

\(\Rightarrow x=100\)

may gioi vai

a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)

b.

\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)

\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)

\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)

\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)

câu 1: Câu hỏi của Vương Ái Như - Toán lớp 7 - Học toán với OnlineMath

câu 2:

Ta có: \(8^7-2^{18}=2^{21}-2^{18}=2^{17}.\left(2^4-2\right)=2^{17}.14⋮14\)

câu 3:

\(4x=7y=3x\Rightarrow\frac{4x}{84}=\frac{7y}{84}=\frac{3z}{84}\Rightarrow\frac{x}{21}=\frac{y}{12}=\frac{z}{28}=\frac{x+y+z}{21+12+28}=\frac{61}{61}=1\)

\(\Rightarrow x=21,y=12,z=28\)

câu 4:

\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)

\(\Rightarrow a=5.12=60,b=9.5=45,c=8.5=40\)

a)\(x-\frac{3}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{3}{5}+\frac{3}{5}=\frac{6}{5}\)

b)\(|x|-\frac{4}{5}=\frac{2}{3}\\ \Rightarrow|x|=\frac{2}{3}+\frac{4}{5}=\frac{22}{15}\\ \Rightarrow|x|=\frac{22}{15}\\ \Rightarrow x=\frac{22}{15}\)

c)\(\frac{x}{-5}=\frac{24}{15}\\ \Rightarrow x=\frac{-5\cdot24}{15}=-8\)

d)\(\frac{x}{4}=\frac{y}{5} và x-y=21\)

Theo tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{x}{4}=\frac{y}{5}=\frac{x-y}{4-5}=\frac{21}{-1}=-21\)

Do đó :

\(\frac{x}{4}=-21\Rightarrow x=-84\)

\(\frac{y}{5}=-21\Rightarrow y=-105\)

\(x-\frac{3}{5}=\frac{3}{5}\)

\(x=\frac{3}{5}+\frac{3}{5}\)

\(x=\frac{6}{5}\)

\(\left|x\right|-\frac{4}{5}=\frac{2}{5}\)

\(\left|x\right|=\frac{2}{5}+\frac{4}{5}\)

\(\left|x\right|=\frac{6}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-\frac{6}{5}\end{cases}}\)

\(\frac{x}{-5}=\frac{24}{15}\)

\(\Rightarrow x.15=\left(-5\right).24\)

\(\Rightarrow x.15=-120\)

\(\Rightarrow x=-120:15\)

\(\Rightarrow x=-8\)

\(\frac{x+1,2}{y}=\frac{11}{5}\Rightarrow\frac{x}{y}+\frac{1,2}{y}=\frac{11}{5}\)

\(\Rightarrow\frac{4}{5}+\frac{1,2}{y}=\frac{11}{5}\Rightarrow\frac{1,2}{y}=\frac{7}{5}\Rightarrow y=1,2:\frac{7}{5}=\frac{6}{7}\)

\(\Rightarrow x=\frac{4}{5}y=\frac{4}{5}.\frac{6}{7}=\frac{24}{35}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)

\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\)\(\ne\)0 nên x + 1 = 0 \(\Rightarrow\)x = -1