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Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
1)
a)
\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)
b)
\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)
c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)
\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)
\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)
\(x\cdot\frac{19}{6}=-\frac{1}{24}\)
x = -1/76
b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
2x = 12
x = 6
a) \(8x-16-21-7x=-4.\) " tính "
" đến đây tim x như bình thường tự làm đi cho não có nếp nhăn "
b) câu B chúa Pain làm nhiều lần rồi đây :)
B1 C/M x không âm
có : \(A=\text{|}x\text{|}+\text{|}x+1\text{|}+\text{|}x+2\text{|}+\text{|}x+3\text{|}\ge0\) " trị tuyệt đối luôn lớn hơn hoặc = 0
mà \(A=6x\) " đề bài "
vậy \(6x\ge0\Leftrightarrow x\ge\frac{0}{6}=0\) vậy x dương
Bước 2 phá trị tuyệt đối với x dương ta được
\(x+x+1+x+2+x+3=6x\)
\(4x+6=6x\)
\(4x-6x=-6\)
\(-2x=-6\Leftrightarrow x=3\)
q, bạn ghi đề rõ nhé
s, \(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{7}{12}=\dfrac{-1}{12}\Leftrightarrow x=-\dfrac{1}{12}:\dfrac{2}{3}=-\dfrac{3}{24}=-\dfrac{1}{8}\)
t, \(\dfrac{3}{4}x=\dfrac{1}{6}-\dfrac{1}{5}=\dfrac{-1}{30}\Leftrightarrow x=-\dfrac{1}{30}:\dfrac{3}{4}=-\dfrac{4}{90}=-\dfrac{2}{45}\)
u, \(\dfrac{1}{6}x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{6}{8}=\dfrac{3}{4}\)
s, 23x=12−712=−112⇔x=−112:23=−324=−1823x=12−712=−112⇔x=−112:23=−324=−18
t, 34x=16−15=−130⇔x=−130:34=−490=−24534x=16−15=−130⇔x=−130:34=−490=−245
u, 16x=38−14=18⇔x=18:16=68=34
HT
a, \(\left(\frac{5}{8}.0,6-5:3\frac{1}{3}\right).\left(\frac{1}{5}-1,4\right).\left(-5\right)^2.x=120\)
\(\left(\frac{5}{8}.\frac{3}{5}-5:\frac{10}{3}\right)\left(\frac{1}{5}-\frac{7}{5}\right).25x=120\)
\(\left(\frac{3}{8}-\frac{3}{2}\right)\left(-\frac{6}{5}\right).25x=120\)
\(\left(-\frac{9}{8}\right).\left(-\frac{6}{5}\right).25x=120\)
\(\frac{27}{20}.25x=120\)\(25x=120:\frac{27}{20}=\frac{800}{9}\)
\(x=\frac{800}{9}:25=\frac{32}{9}\)
đợi tí đi nấu cơm đã
\(\Leftrightarrow2^{6x-1-2}=2^9\)
=>6x-3=9
=>6x=12
=>x=2
`@` `\text {Answer}`
`\downarrow`
`h)`
\(2^{6x-1}\div4=8^3\)
`=>`\(2^{6x-1}=\left(2^3\right)^3\cdot4\)
`=>`\(2^{6x-1}=2^9\cdot2^2\)
`=>`\(2^{6x-1}=2^{11}\)
`=> 6x - 1 = 11`
`=> 6x = 11 +1`
`=> 6x = 12`
`=> x = 12 \div 6`
`=> x = 2`
Vậy, `x = 2.`