tìm x ∈ N, biết:
\(\frac{1}{4}\)+ \(\frac{8}{9}\)≤ \(\frac{x}{36}\)< 1- (\(\frac{3}{8}\)- \(\frac{5}{6}\))
tìm x thuộc N,biết:\(\frac{1}{4}+\frac{8}{9}\)< hoặc bằng \(\frac{x}{36}< 1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}\le1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
<=> \(\frac{41}{36}\le\frac{x}{36}\le\frac{35}{24}\)
<=> \(\frac{82}{72}\le\frac{2x}{72}\le\frac{105}{72}\)
<=> \(82\le2x\le105\)
<=> \(41\le x\le52,5\)
Do \(x\in N\)nên \(x=\left\{x\in N|41\le x\le52,5\right\}\)
Giúp với, tìm các số nguyên x biết:
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}< 1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}< 1-\frac{3}{8}+\frac{5}{6}\)
tìm x ; x e Z
C =22−3 x 4− x =12−3 x +104− x =3(4− x )4− x +104− x =3+104− x ... +199.100 ). x =201251 +201252 +. .... =12√3 (4+2√3√3+1+4−2√3√3−1 )=12√3 .4√3−4+6−2√3+4√3+4−6−2√3(√3−1)(√3+1).
\(\frac{1}{4}+\frac{8}{9}< \frac{x}{36}< 1-\frac{3}{8}+\frac{5}{6}\)
\(=\frac{41}{36}< \frac{x}{36}< \frac{35}{24}\)
=>\(x=35< x< 41\)
Tìm x biết: \(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}.\)
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Tìm x
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{9}{10}.....\frac{36}{62}.\frac{31}{64}=\frac{1}{2^x}\)
1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x
(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x
(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x
(31 số 1/2)
(1/2)^31. = 2^x
=> 0=x+36
x=0-36
x=-36
Vậy x=-36
Theo mk nghĩ,mk làm đúng nha .Tk cho mk
Để mk sửa phần này một chút
\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)
\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)
\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)
\(\frac{1}{2^{36}}=2^x\)
\(1=2^x\cdot2^{36}\)
\(2^0=2^x+36\)
Rồi bn tự suy luận nha
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.........................\frac{31}{64}=\frac{1}{2^x}\)
Có 31 thừa số
\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}\right)...................\left(\frac{1}{2}.\frac{31}{32}\right)=\frac{1}{2^x}\)
\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}...............\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}...............\frac{31}{32}\right)=\frac{1}{2^x}\)
Có 31 thừa số Có 31 thừa số
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1.2....................31}{2.3..............32}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{32}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{2^5}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{36}}=\frac{1}{2^x}\)
\(\Leftrightarrow x=36\)
Vậy x=36
Chúc bạn học tốt
a, tính:\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b, tìm x biết:\(1\frac{1}{30}:(24\frac{1}{6}-24\frac{1}{5})-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=(-1\frac{1}{5}):(8\frac{1}{5}-8\frac{1}{3})\)
bài 15 tìm x biết
a\(\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)
b\(\frac{5}{12}+\frac{5}{x}-\frac{1}{8}=\frac{1}{2}\)
c\(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
d\(\frac{3}{4}-x+\frac{6}{-11}=\frac{5}{6}\)
e\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
tìm X biết
\(a,\frac{3}{5}-x=0.2;b,\frac{x}{3}-\frac{1}{8}=\frac{5}{8};c,3\frac{1}{3}x-6\frac{3}{4}=3\frac{1}{4}\)
tìm x biết
1, \(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
2, \(\frac{x}{120}=\frac{3}{8}\cdot\frac{-4}{25}\)
3, \(\frac{11}{13}\cdot x=1\)
4, \(-\frac{9}{8}+\frac{-3}{8}\cdot x=\frac{-1}{8}\)
5, \(\frac{-2}{5}\cdot x+\frac{4}{3}=\frac{7}{3}\)
\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
\(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)
\(x-\frac{3}{5}=\frac{263}{210}\)
\(x=\frac{263}{210}+\frac{3}{5}\)
\(x=\frac{389}{210}\)
VẬY: \(x=\frac{389}{210}\)