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Câu hỏi của SKT_NTT - Toán lớp 5 - Học toán với OnlineMath
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Xét vế phải
\(\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\frac{9}{1}\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+9\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\left(1+1+...+1\right)\)
\(=\left(1+\frac{1}{9}\right)+\left(1+\frac{2}{8}\right)+\left(1+\frac{3}{7}\right)+\left(1+\frac{4}{6}\right)+\left(1+\frac{5}{5}\right)+\left(1+\frac{6}{4}\right)+\left(1+\frac{7}{3}\right)+\left(1+\frac{8}{2}\right)+1\)\(=\frac{10}{9}+\frac{10}{8}+\frac{10}{7}+\frac{10}{6}+\frac{10}{5}+\frac{10}{4}+\frac{10}{3}+\frac{10}{2}+\frac{10}{10}\)
\(=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Thay vào bài ,ta được:
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\times x=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)\(\Rightarrow x=10\)
Vậy x=10
chúc bạn học tốt
\(\frac{4}{5}-\left|3-x\right|=\frac{1}{4}\)
\(\Rightarrow\left|3-x\right|=\frac{4}{5}-\frac{1}{4}\)
\(\Rightarrow\left|3-x\right|=\frac{11}{20}\)
\(\Rightarrow\hept{\begin{cases}3-x=\frac{11}{20}\\3-x=-\frac{11}{20}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{49}{20}\\x=\frac{71}{20}\end{cases}}\)
\(\frac{8}{9}:x^2+\frac{1}{9}=1\)
\(\Rightarrow\frac{8}{9}:x^2+\frac{1}{9}=\frac{9}{9}\)
\(\Rightarrow\frac{8}{9}:x^2=\frac{9}{9}-\frac{1}{9}\)
\(\Rightarrow\frac{8}{9}:x^2=\frac{8}{9}\)
\(\Rightarrow x^2=\frac{8}{9}:\frac{8}{9}\)
\(\Rightarrow x^2=1\)
\(\Rightarrow x^2=1^2\)
\(\Rightarrow x=1\)
Thêm:
\(\frac{4}{5}+\frac{3}{4}.x\frac{5}{2}\)
\(x-\frac{1}{2}=-\frac{5}{2}\)
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\left(x-1\right)\cdot3=8\cdot9\)
\(\left(x-1\right)\cdot3=72\)
\(x-1=\frac{72}{3}\)
\(x-1=24\)
\(x=24+1\)
\(x=25\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(x\cdot\left(x+1\right)=18\cdot4\)
\(x\left(x+1\right)=72\)
\(x\left(x+1\right)=8\cdot9\)
\(x=8\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\frac{x}{-4}=\frac{-9}{x}\)
\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)
\(x^2=36\)
\(x^2=\left(-6\right)^2\)hoặc \(x^2=6^2\)
\(x=-6\) hoặc\(x=6\)
a)x-1/9=24/9 => x-1=24 =>x=23
b)x(x+1)=18*4 =>x=8
c)-x:4=-9:x =>-1.x2=-1.36 =>x=6
k mik nha!
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
\(\frac{x+1}{9}+\frac{x+2}{8}+\frac{x+3}{7}+...+\frac{x+9}{1}=-9\)
\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)+\left(\frac{x+3}{7}+1\right)+...+\left(\frac{x+9}{1}+1\right)=0\)
\(\frac{x+10}{9}+\frac{x+10}{8}+\frac{x+10}{7}+...+\frac{x+10}{1}=0\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\right)=0\)
vì \(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\ne0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
\(\frac{3}{9}\)là đáp án
4/9-x/x=1/9
x/x=4/9-1/9
x/x=1/3
k mk nha!