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ừ Vy Nguyễn, mik làm nè:
e, \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-4}{6}+\dfrac{-9}{6}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}.\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}.\)
\(2x-5=\dfrac{-13}{6}.3.\)
\(2x-5=\dfrac{-13}{2}.\)
\(2x=\dfrac{-13}{2}+5.\)
\(2x=\dfrac{-13}{2}+\dfrac{10}{2}.\)
\(2x=\dfrac{-3}{2}.\)
\(x=\dfrac{-3}{2}:2.\)
\(x=\dfrac{-3}{2.2}=\dfrac{-3}{4}.\)
g, \(\dfrac{2}{5}x+\dfrac{1}{2}=\dfrac{-3}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}-\dfrac{1}{2}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}+\dfrac{-2}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-5}{4}.\)
\(x=\dfrac{-5}{4}:\dfrac{2}{5}.\)
\(x=\dfrac{-5}{4}.\dfrac{5}{2}.\)
\(x=\dfrac{-25}{8}.\)
h, \(\left(2x-2\dfrac{4}{5}\right):3\dfrac{1}{8}=1\dfrac{3}{5}.\)
\(\left(2x-2\dfrac{4}{5}\right)=\dfrac{8}{5}.\dfrac{25}{8}.\)
\(\left(2x-2\dfrac{4}{5}\right)=5.\)
\(2x=5+2\dfrac{4}{5}.\)
\(2x=7\dfrac{4}{5}.\)
\(x=7\dfrac{4}{5}:2.\)
\(x=\dfrac{39}{10}.\)
(còn tiếp ở phần sau!!!)
Tiếp:
i, \(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{7}{20}.\dfrac{11}{3}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\dfrac{22}{15}=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{22}{15}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{88}{60}.\)
\(\dfrac{16}{5}x=\dfrac{165}{60}=\dfrac{11}{4}.\)
\(x=\dfrac{11}{4}:\dfrac{16}{5}.\)
\(x=\dfrac{11}{4}.\dfrac{5}{16}=\dfrac{55}{64}.\)
k, \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}.\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right).\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}.\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}.\)
\(\Rightarrow3x=-6.\)
\(\Rightarrow x=-6:3=-2.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tik mik nha!!!
\(1,3.\dfrac{15}{39}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(=\dfrac{13}{10}.\dfrac{15}{39}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\)
\(4\dfrac{1}{3}\cdot\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\\ \dfrac{13}{3}\cdot\dfrac{-1}{3}\le x\le\dfrac{2}{3}\cdot\dfrac{-11}{12}\\ \dfrac{-13}{9}\le x\le\dfrac{-11}{18}\\ \dfrac{-26}{18}\le x\le\dfrac{-11}{18}\\ \Rightarrow x=-1\)
\(x:4\dfrac{1}{3}=2,5\\ x:\dfrac{13}{3}=\dfrac{5}{2}\\ x=\dfrac{5}{2}.\dfrac{13}{3}\\ x=\dfrac{65}{6}=10\dfrac{5}{6}\)
Mik làm luôn mik ko chép đề đâu
1)
a) \(\left|x\right|=\dfrac{5}{9}+\dfrac{3}{5}\)
\(\left|x\right|=\dfrac{25}{45}+\dfrac{27}{45}\)
\(\left|x\right|=\dfrac{52}{45}\)
\(\Rightarrow x=\dfrac{52}{45}or\left(-\dfrac{52}{45}\right)\)
Mà x>0
\(\Rightarrow x=\dfrac{52}{45}\)
Vậy \(x=\dfrac{52}{45}\)
b) \(-2\left|x\right|=\dfrac{-4}{3}\)
\(\left|x\right|=\dfrac{-4}{3}:\left(-2\right)\)
\(\left|x\right|=\dfrac{-4}{3}.\dfrac{-1}{2}\)
\(\left|x\right|=\dfrac{4}{6}\)
\(\left|x\right|=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{3}or\left(-\dfrac{2}{3}\right)\)
Mà x<0
\(\Rightarrow x=-\dfrac{2}{3}\)
a) \(\dfrac{x+3}{x-5}=\dfrac{x-5+8}{x-5}=\dfrac{x-5}{x-5}+\dfrac{8}{x-5}=1+\dfrac{8}{x-5}\)
Để \(\dfrac{x+3}{x-5}\) có giá trị âm thì \(8⋮x-5\) và \(x-5< 0\)
\(\Rightarrow x-5\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Để \(x-5< 0\Rightarrow x< 5\)
Nên \(x\in\left\{\pm1;\pm2;\pm4;-8\right\}\)
~ Học tốt ~
1) Tìm x
a) B(32) = { 0 , 32 , 64 , 96 , 128 ; 160 ; 192 ; ... }
b) \(\dfrac{11}{12}\) - ( \(\dfrac{2}{5}\) +x ) = \(\dfrac{2}{3}\)
\(\dfrac{2}{5}\) + x = \(\dfrac{11}{12}\) - \(\dfrac{2}{3}\)
\(\dfrac{2}{5}\) +x = \(\dfrac{1}{4}\)
x = \(\dfrac{1}{4}\) - \(\dfrac{2}{5}\)
x = \(\dfrac{-3}{20}\)
B(41 ) = { 0 , 41 , 82 , 123 , 164 , 205 , .... }
c ) 2.( 2x-\(\dfrac{1}{7}\) ) = 0
=> \(2\text{x}-\dfrac{1}{7}\) = 0
=> 2x = \(\dfrac{1}{7}\)
=> x = \(\dfrac{1}{14}\)
d) ( 3 - 2x ) (7x - \(\dfrac{1}{8}\) ) = 0
=> 3-2x = 0 hoặc 7x - \(\dfrac{1}{8}\) =0
* Nếu 3 - 2x = 0
=> 2x = 3
=> x = \(\dfrac{3}{2}\)
*Nếu 7x - \(\dfrac{1}{8}\) = 0
=> 7x = \(\dfrac{1}{8}\)
=> x = \(\dfrac{1}{56}\)
Vậy x = \(\dfrac{3}{2}\) hoặc x = \(\dfrac{1}{56}\)
2) Xác định giá trị của x để :
a) \(\dfrac{x+3}{x-5}\) có giá trị âm
=> x+3 phải là số nguyên dương
=> x-5 phải là số nguyên âm
b) Để ( \(x+\dfrac{2}{3}\) ) . ( x - 2 ) > 0
=> ( \(x+\dfrac{2}{3}\) ) và ( x-2 ) \(\in\) N*