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A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
\(\frac{1}{2}+\frac{5}{6}-\frac{3}{8}\)
\(=\frac{12}{24}+\frac{20}{24}-\frac{9}{24}\)
\(=\frac{23}{24}\)
\(\frac{10}{15}\cdot\frac{7}{4}+\frac{7}{4}\cdot\frac{9}{15}-\frac{4}{15}\cdot\frac{7}{4}\)
\(=\frac{7}{4}\cdot\left(\frac{10}{15}+\frac{9}{15}-\frac{4}{15}\right)\)
\(=\frac{7}{4}\cdot1=\frac{7}{4}\)
\(\left(\frac{4}{5}+\frac{1}{2}\right)\left(\frac{3}{13}-\frac{8}{13}\right)\)
\(=\left(\frac{8}{10}+\frac{5}{10}\right)\cdot\left(-\frac{5}{13}\right)\)
\(=\frac{13}{10}\cdot\left(-\frac{5}{13}\right)=-\frac{1}{2}\)
a, 3x-12 = 30
=> 3x = 30 + 12
=> 3x = 42
=> x = 42 : 3 = 14
Vậy x = 14
b, \(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}\div\frac{2}{3}\Rightarrow\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
c, 2x2 = 32
=> x2 = 32 : 2
=> x2 = 16
=> x2 = 42
=> x = 4
Vậy x = 4
\(a.\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90\cdot\frac{2}{3}=-30\cdot\frac{2}{1}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=28\)
\(x=\frac{28}{2,8}=10\). Vậy x=10
\(b.\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{14}\)
\(4,5-2x=\frac{11}{14}:\frac{11}{7}=\frac{11}{14}\cdot\frac{7}{11}\)
\(4,5-2x=\frac{1}{2}\)
\(2x=\frac{45}{10}-\frac{1}{2}=\frac{9}{2}-\frac{1}{2}=\frac{8}{2}\)
\(2x=4\)
\(x=\frac{4}{2}=2\) . Vậy x=2
Chúc bạn học tốt!^_^
a)(2,8x-32):\(\frac{2}{3}\)=-90
(2,8x-32) =(-90).\(\frac{2}{3}\)
(2,8x-32) =(-60)
2,8x =(-60)+32
2,8x =(-28)
x =(-28):2,8
x =(-10)
b)(4,5-2x).1\(\frac{4}{7}\)=\(\frac{11}{14}\)
(4,5-2x).\(\frac{11}{7}\)=\(\frac{11}{14}\)
(4,5-2x) =\(\frac{11}{14}\):\(\frac{11}{7}\)
(4,5-2x) =\(\frac{1}{2}\)
2x =4,5-\(\frac{1}{2}\)
2x =4
x =4:2
x =2
a)(2,8x-32):\(\frac{2}{3}\)=-90
2,8x-32=-90.\(\frac{2}{3}\)
2,8x =-60+32
x =-28:2,8
x=-10
b)(4,5-2x).1\(\frac{4}{7}\)=\(\frac{11}{14}\)
4,5-2x =\(\frac{11}{14}:\frac{11}{7}\)
-2x =\(\frac{1}{2}\)
x=-1/4
d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)
=> \(2x+7=4\)
=> 2x= -3
=> x=-3/2 . Vậy x=-3/2
e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)
=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)
= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=> \(7^x=5^{2x}\)
Đến đoạn này là mik nghĩ không ra nhé
Cô làm tiếp giúp Linh Đan:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
2x-1+5.2x-2=2.2x-2+5.2x-2=2x-2.(5+2)=2x-2.7=7/32
=>2x-2=7/32:7=1/32
=>x-2=-5
=>x=-3
ĐOán thế
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