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\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)

a) Ta có: \(-x+\frac{4}{7}=\frac{1}{3}\)
\(\Leftrightarrow-x=-\frac{5}{21}\)
\(\Rightarrow x=\frac{5}{21}\)
b) Ta có: \(x\div\left(-\frac{1}{3}\right)^2=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^3=-\frac{1}{27}\)
c) \(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{5}\right)^2=\frac{9}{25}\)
\(a.-x+\frac{4}{7}=\frac{1}{3}\)
\(-x=\frac{1}{3}-\frac{4}{7} \)
\(-x=\frac{7}{21}-\frac{12}{21}\)
\(-x=\frac{-5}{21}\)
\(x=\frac{5}{21}\)
\(b.x:\left(\frac{-1}{3}\right)^2=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\left(\frac{-1}{3}\right)^2\)
\(x=\frac{-1}{3}.\frac{-1}{3}.\frac{-1}{3}\)
\(x=\frac{-1}{27}\)
\(c.\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(x=\left(\frac{3}{5}\right)^7:\left(\frac{3}{5}\right)^5\)
\(x=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}.\frac{3}{5}\)
\(x=\frac{9}{25}\)

1.\(\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)
\(=\frac{5}{7}.\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}.\frac{-7}{11}=\frac{5.\left(-7\right)}{7.11}=\frac{5.\left(-1\right)}{1.11}=\frac{-5}{11}\)
\(C=\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)
\(=\frac{-3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(=\frac{-3}{7}.1+2\frac{3}{7}=\frac{-3}{7}+2\frac{3}{7}=2\)

\((2,7.x-1\frac{1}{2})\div\frac{2}{7}=\frac{-21}{4}\) \(3\frac{1}{3}.x+16\frac{3}{4}=-13.25\)
\(2,7.x-1\frac{1}{2}=-\frac{21}{4}\cdot\frac{2}{7}\) \(\frac{10}{3}.x+\frac{67}{4}=-13.25\)
\(2,7.x-\frac{3}{2}=-\frac{3}{2}\) \(\frac{10}{3}.x+\frac{67}{4}=-\frac{53}{4}\)
\(2,7.x=-\frac{3}{2}+\frac{3}{2}\) \(\frac{10}{3}.x=-\frac{53}{4}-\frac{67}{4}\)
\(2,7.x=0\) \(\frac{10}{3}.x=-30\)
\(x=0:2,7\) \(x=-30:\frac{10}{3}\)
\(x=0\) \(x=-9\)
Vậy x=0 Vậy x= -9
\(\left(4.5-2.x\right):\frac{3}{4}=1\frac{1}{3}\) \(1.5+1\frac{1}{4}.x=\frac{2}{3}\)
\(\left(4.5-2.x\right)=1\frac{1}{3}\cdot\frac{3}{4}\) \(1\frac{1}{4}.x=\frac{2}{3}-1.5\)
\(4.5-2.x=\frac{4}{3}\cdot\frac{3}{4}\) \(\frac{5}{4}.x=\frac{2}{3}-\frac{3}{2}\)
\(4.5-2.x=1\) \(\frac{5}{4}.x=-\frac{5}{6}\)
\(2.x=4.5-1\) \(x=-\frac{5}{6}:\frac{5}{4}\)
\(2.x=3.5\) \(x=-\frac{2}{3}\)
\(x=3.5:2\)
\(x=1.75\) Vậy \(x=-\frac{2}{3}\)
Vậy x=1.75

Ta có : \(\left(5x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\2x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=3\\2x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{5}{2}\end{cases}}\)

a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d

a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)

a)(1,5.x+3/7):1,5-1,5=1 b)(2+x+4/7+3/7):1,4-5/7=-2
1,5.x:1,5+3/7:3/2=1+1,5 (2+x+1):1,4-5/7=-2
x+ 3/7x 2/3=2,5 (3+x):1,4=-2+5
x+2/7=2,5 (3+x):1,4=3
x+2/7=5/2 3+x=3x1,4
x=5/2-2/7 3+x=4,2
x=31/14 x=4,2-3=1,2
cái này thì làm j bn,ko có câu hỏi à?
X ở dou bn
có x ở đâu mà tính
sửa lại đề đi mk giải cho nha
nhanh lên