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Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
bài 2
a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)
b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)
bài 1
a] y là 9
b] <=> 64y + 36y = 700 - 75 - 225
<=> 100y = 400
<=> y = 4
trên lớp cô sửa rồi nên mình giải luôn:
1) Tìm y
a) y3 + 3y = 12 x 11
y3 + 3y = 132
y x 10 + 3 + 3 x 10 + y = 132
( y x 10 + y ) + ( 3 x 10 + 3 ) = 132
11 x y + 33 = 132
11 x y = 132 - 33
11 x y = 99
y = 99 : 11
y = 9
b) 64 x y + 225 = 700 - 75 - 36 x y
64 x y + 225 = 625 - 36 x y
64 x y + 36 x y = 625 -225
64 x y + 36 x y = 400
( 64 + 36 ) x y = 400
100 x y = 400
y = 400 : 100
y = 4
2) Tính
a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
\(=\frac{4343}{7171}\times4\)
\(=\frac{43}{71}\times4\)
\(=\frac{172}{71}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Ta có:
\(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)
\(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)
\(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)
A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)
A = 1 x \(\frac{4}{21}\)
A = \(\frac{4}{21}\)
a)\(\frac{x}{17}=\frac{60}{204}=\frac{5}{17}\Rightarrow x=5\)
b)\(\frac{6+x}{33}=\frac{7}{11}\Rightarrow11\left(6+x\right)=7.33\Rightarrow11.6+11x=231\Rightarrow66+11x=231\)
\(\Rightarrow11x=231-66\Rightarrow11x=165\Rightarrow x=\frac{165}{11}=15\)
c)\(\frac{12+x}{43-x}=\frac{2}{3}\Rightarrow2\left(43-x\right)=3\left(12+x\right)\Rightarrow2.43-2x=3.12+3x\)
\(86-2x=36+3x\Rightarrow86-36=3x+2x\Rightarrow50=5x\Rightarrow x=\frac{50}{5}=10\)
a,
\(x+\frac{7}{15}=1\frac{1}{20}\)
\(x+\frac{7}{15}=\frac{21}{20}\)
\(x=\frac{21}{20}-\frac{7}{15}=\frac{63}{60}-\frac{28}{60}\)
\(x=\frac{35}{60}=\frac{7}{12}\)
b,
\(\left[3\frac{1}{2}-x\right]\cdot1\frac{1}{4}=\frac{15}{16}\)
\(\left[\frac{7}{2}-x\right]\cdot\frac{5}{4}=\frac{15}{16}\)
\(\frac{7}{2}-x=\frac{15}{16}:\frac{5}{4}=\frac{3}{4}\)
\(\frac{7}{2}-x=\frac{3}{4}\Rightarrow x=\frac{7}{2}-\frac{3}{4}\)
\(x=\frac{11}{4}\)
c,
\(1\frac{1}{5}x+\frac{2}{3}x=\frac{56}{125}\Leftrightarrow\frac{6}{5}x+\frac{2}{3}x=\frac{56}{125}\)
\(\frac{28}{15}x=\frac{56}{125}\Rightarrow x=\frac{6}{25}\)
d,
\(60\%x+0,4x+x:3=2\)
\(\frac{3}{5}x+\frac{2}{5}x+\frac{1}{3}x=2\)
\(\frac{4}{3}x=2\Rightarrow x=\frac{3}{2}\)
Nguyễn Anh Thiện
a)
x + \(\frac{7}{15}\) = \(1\frac{1}{20}\)
X + \(\frac{7}{15}=\frac{21}{20}\)
X \(=\frac{21}{20}-\frac{7}{15}\)
X \(=\frac{63}{60}-\frac{28}{60}=\frac{35}{60}=\frac{7}{12}\)
^^ Học tốt !