\(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{2017}}\)
B=1+2+22+23+.............+22017
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em c.ơn trước
Y
9 tháng 2 2023
\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)
\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)
\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow x=-1\left(loai\right)\)
Vậy \(S=\varnothing\)
NM
1
17 tháng 8 2017
\(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow\left(x-2\right)3=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=-6\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy .....
b, \(B=1+2+2^2+..........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=3x+120\)
\(\Leftrightarrow4x-3x=120-92\)
\(\Leftrightarrow x=28\)
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
\(21x9=189\left(cm2\right)\)
24 tháng 9 2022
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)
=>-10<=x<=-13/7
hay \(x\in\left\{-10;-9;...;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=11/18
hay \(x\in\left\{-1;0\right\}\)
TD
13 tháng 2 2018
a:
\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)
\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)
\(\Leftrightarrow0x=88\)
Vậy pt vô nghiệm
b:
\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)
\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)
\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)
\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)
\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)
\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)
Vậy tập nghiện của pt S= {-38}
c:
\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)
\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)
\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)
\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)
\(\Leftrightarrow0x=0\)
Vậy pt vô số nghiệm
BC
16 tháng 3 2022
Bài 3
\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)
\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)
\(=\dfrac{33}{23}\le x\le3\)
Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)
Có 1 giá trị thỏa mãn
Chọn A
Bài 4
\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)
Chọn D
Bài 5
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{100}{100}-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
CHọn C
TT
12 tháng 5 2021
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
21 tháng 5 2022
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow3\left(x-2\right)=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=2x+80\)
\(\Leftrightarrow4x-2x=80-92\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)
d, \(B=1+2+2^2+........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)
\(< =>3x-6=5x=>x=-3\)
\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)
\(4x+92=3x+120=>x=28\)