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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*
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
tính
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
MONG CÁC BẠN GIÚP ĐỠ XIN CẢM ƠN
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\)
\(=\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{13}{56}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\\ =\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\left(\dfrac{5}{8}\cdot\dfrac{1}{5}\right)-\left(\dfrac{3}{16}\cdot4\right)\\ =\dfrac{6}{7}+\left(\dfrac{1\cdot1}{8\cdot1}\right)-\left(\dfrac{3}{4}\cdot1\right)\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{42}{56}\\ =\dfrac{48+7-42}{56}\\ =\dfrac{13}{56}\)
a)\(\dfrac{-1}{4}\cdot13\dfrac{9}{11}-0,25\cdot6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
=\(\dfrac{1}{4}\cdot\left(\dfrac{-152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=-5\)
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!!!
a, (3\(\dfrac{1}{5}-2\dfrac{1}{3}+\dfrac{5}{6}\)) - (1,5 - \(\dfrac{2}{3}+\dfrac{1}{6}\))
= \(\dfrac{16}{5}-\dfrac{7}{3}+\dfrac{5}{6}-\dfrac{3}{2}+\dfrac{2}{3}-\dfrac{1}{6}\)
= (\(\dfrac{16}{5}-\dfrac{3}{2}\)) - (\(\dfrac{7}{3}-\dfrac{2}{3}\)) + (\(\dfrac{5}{6}-\dfrac{1}{6}\))
= \(\dfrac{17}{10}-\dfrac{5}{3}+\dfrac{2}{3}\)
= \(\dfrac{17}{10}-1\)
= \(\dfrac{7}{10}\)
@Christina
b, \(\dfrac{\left(3\dfrac{1}{5}.1,4-2,5.\dfrac{7}{108}\right):2\dfrac{7}{18}+4\dfrac{1}{2}.0,1}{70,5-5,28:7\dfrac{1}{2}}\)
= \(\dfrac{\left(\dfrac{16}{5}.\dfrac{7}{5}-\dfrac{5}{2}.\dfrac{7}{108}\right):\dfrac{43}{18}+\dfrac{9}{2}.\dfrac{1}{10}}{\dfrac{141}{2}-\dfrac{132}{25}:\dfrac{15}{2}}\)
= \(\dfrac{\left(\dfrac{112}{25}-\dfrac{35}{216}\right):\dfrac{43}{18}+\dfrac{9}{20}}{\dfrac{141}{2}-\dfrac{88}{125}}\)
= \(\dfrac{\dfrac{112}{25}:\dfrac{43}{18}-\dfrac{25}{216}:\dfrac{43}{18}+\dfrac{9}{20}}{\dfrac{17449}{250}}\)
= \(\dfrac{\dfrac{2016}{1075}-\dfrac{25}{516}+\dfrac{9}{20}}{\dfrac{17449}{250}}\)
= \(\dfrac{\dfrac{2016}{1075}+\dfrac{259}{645}}{\dfrac{17449}{250}}\)
= \(\dfrac{\dfrac{7343}{3225}}{\dfrac{17449}{250}}\)
@Christina
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
\(\Rightarrow x=-\dfrac{2}{3}\)