a)\(\dfrac{1}{18}\)< \(\dfrac{x}{12}\)<\(\dfrac{y}{9}\)<\(\dfrac{1}{4}\) ; b)\(\dfrac{7}{8}\)< \(\dfrac{x}{40}\)< \(\dfrac{9}{10}\)
mk cần gấp lắm,ai nhanh mk like cho
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Những câu hỏi liên quan
17 tháng 4 2021
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
17 tháng 4 2021
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
11 tháng 8 2021
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
VQ
7 tháng 5 2022
bài 1 :
\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)
\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)
\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)
\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)
bài 2 :
\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}-\dfrac{1}{3}\)
\(x=\dfrac{1}{12}\)
\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}x\dfrac{7}{4}\)
\(x=\dfrac{7}{10}\)
bài 3 :
đổi : 2 dm 1cm = 21cm
chiều cao hình bình hành là;
21 x\(\dfrac{3}{7}=\)9(cm)
diện tích hình bình hành là;
21 x 9 =189 (cm2)
đáp số : 189 cm2
bài 4 :
\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)
\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)
=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)
\(=\dfrac{2}{3}x\dfrac{2}{3}\)
=\(\dfrac{4}{9}\)
C
7 tháng 5 2022
Bài 1)
a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)
b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)
c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)
d) \(\dfrac{6}{414}=\dfrac{1}{69}\)
Bài 2)
a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)
\(x=\dfrac{1}{12}\)
b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)
\(x=\dfrac{7}{10}\)
Bài 3)
2dm 1cm = 21 cm
Chiều cao tấm bìa la
\(21x\dfrac{3}{7}=9\left(cm\right)\)
Diện tích tấm bìa là
\(21x9=189\left(cm2\right)\)
28 tháng 12 2021
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
VG
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 7 2023
\(\dfrac{5}{12}\)\(x\) - \(\dfrac{1}{4}\)\(x\) = - \(\dfrac{13}{18}\)
(\(\dfrac{5}{12}\) - \(\dfrac{1}{4}\))\(x\) = - \(\dfrac{13}{18}\)
\(\dfrac{1}{6}\)\(x\) = - \(\dfrac{13}{18}\)
\(x\) = - \(\dfrac{13}{18}\)\(\times\) 6
\(x\)= -\(\dfrac{13}{3}\)
TV
3
28 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\)
=>(x+12)(x-3)=0
=>x=-12 hoặc x=3
28 tháng 2 2022
\(ĐKXĐ:x\ne-3,-4,-5,-6\)
\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{3}{x^2+9x+18}=\dfrac{1}{18}\\ \Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-12\left(tm\right)\end{matrix}\right.\)
GL
24 tháng 8 2023
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
Q
2
30 tháng 6 2021
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
a) ta co:
1/18<x/12<y/9<1/4
=>2/36<x.3/36<y.4/36<9/36
=>x.3thuộc{3;6};y.4thuộc{4;8}
=>x thuộc{1;2};y thuộc{1:2}
b) ta co
7/8<x/40<9/10
=>70/80<x.2/40<72/80
=>x.2 =71
=>x=71/2