a) \(\dfrac{2}{15}\cdot x-x=30\%\)
b) \(\dfrac{x}{2}\cdot\dfrac{x}{3}=\dfrac{5}{6}\)
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AH
Akai Haruma
Giáo viên
15 tháng 11 2017
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)
Xét mẫu số:
\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)
\(=2^{32}(1.2.3....31.32)\)
Suy ra:
\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)
Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)
\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)
Vậy \(x=\frac{-37}{2}\)
YN
21 tháng 6 2022
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
T
13 tháng 7 2022
a)\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}(21+1,5)⋅x=51
2 \cdot x=\dfrac{1}{5}2⋅x=51
x=\dfrac{1}{5}: 2x=51:2
x=\dfrac{1}{10} x=101
b) \left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}(−153+x):1312=261
-1 \dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}−153+x=613⋅1312
x=2+1 \dfrac{3}{5}x=2+153
x=3 \dfrac{3}{5} x=353
c) \left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}(x:231)⋅71=8−3
x \cdot \dfrac{3}{7} \cdot \dfrac{1}{7}=\dfrac{-3}{8}x⋅73⋅71=8−3
x=\dfrac{-3}{8}: \dfrac{3}{49}x=8−3:493
x=\dfrac{-49}{8}=-6 \dfrac{1}{8}x=8−49=−681
d) \dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)7−4⋅x+57=81:(−132)
\dfrac{-4}{7} x+\dfrac{7}{5}=\dfrac{1}{8} \cdot \dfrac{-3}{5}7−4x+57=81⋅5−3
-\dfrac{4}{7} x=\dfrac{-3}{40}-\dfrac{7}{5} \\ x=\dfrac{-59}{40}: \dfrac{-4}{7}=\dfrac{413}{160}=2 \dfrac{93}{160}−74x=40−3−57x=40−59:7−4=160413=216093
20 tháng 6 2018
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
G
2 tháng 8 2017
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
Chúc bạn học tốt!
NA
2 tháng 8 2017
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NHỚ TRẢ LỜI NHANH GIÙM MK NHA
NA
2 tháng 8 2017
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2 tháng 8 2017
a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)
\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)
\(\Rightarrow x=\dfrac{-79}{210}\)
b) Tương tự câu a)
H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)
\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)
\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)
b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)
\(=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4}{x\left(x+3\right)}\)
c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}\)
\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)
\(=\dfrac{2\left(a-3\right)}{a}\)
d) \(\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)\)
\(=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}\)
\(=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}\)
\(=x+\dfrac{2-x^2}{x\left(x-1\right)}\)
14 tháng 1 2023
a: \(\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^2}=\dfrac{1}{x}\)
=>x^2=x(x^2-x+1)
=>x(x-x^2+x-1)=0
=>x(-x^2+2x-1)=0
=>x=0(loại) hoặc x=1(nhận)
b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2
=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)
=>\(x\simeq28,4\)
a)\(\dfrac{2}{15}x-x=30\%\)
<=>x\(\left(\dfrac{2}{15}-1\right)=\dfrac{30}{100}\)
<=>\(-\dfrac{13}{15}x=\dfrac{3}{10}\)
<=>x=\(\dfrac{3}{10}\cdot\left(-\dfrac{15}{13}\right)\)
<=>x=\(-\dfrac{9}{26}\)
Vậy tập nghiệm S={\(-\dfrac{9}{26}\)}
b)\(\dfrac{x}{2}\cdot\dfrac{x}{3}=\dfrac{5}{6}\)
<=>\(\dfrac{x^2}{6}=\dfrac{5}{6}\)
<=>x2=5
<=>x=\(_-^+\sqrt{5}\)
Vậy tập nghiệm S={\(_-^+\sqrt{5}\)}
a, \(\dfrac{2}{15}x-x=30\%\)
\(\Rightarrow\left(\dfrac{2}{15}-1\right)x=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{-13}{15}x=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}.\dfrac{-15}{13}\)
\(\Rightarrow x=\dfrac{-9}{26}\)Lời giải:
b, \(\dfrac{x}{2}.\dfrac{x}{3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x.x}{2.3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x^2}{6}=\dfrac{5}{6}\)
\(\Rightarrow x^2=5\)
\(\Rightarrow x=\pm\sqrt{5}\)