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\(\frac{7}{3}x+x=42\)
\(\frac{7}{3}x+1x=42\)
\(\left(\frac{7}{3}+1\right)x=42\)
\(\frac{10}{3}x=42\)
\(x=42:\frac{10}{3}\)
\(x=12,6\)
\(\frac{7}{3}.x+x=42\)
\(\Rightarrow x.\left(\frac{7}{3}+1\right)=42\)
\(\Rightarrow x.\frac{10}{3}=42\)
\(\Rightarrow x=8,4\)
Vậy x = 8,4
B) 3/7 . x + x = 42
x(3/7+1) = 42
10/7 . x = 42
x= 42.7/10
x= 147/5
Ta có :
\(\text{A = 1.2.3 + 3.4.5+...99.100.101}\)
\(\text{A=1.3(5-3)+3.5(7-3)+}...+99.101\left(103-3\right)\)
\(=\left(1.3.5+3.5.7+5.7.9+...99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)
\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+...+99.101\right)\)
\(=13517400-3.171650\)
\(=13002450\)
D=1.2.3+3.4.5+...+99.100.101
D=1.2.3.4+5.6.7.4+........+99.100.101.4
D=1.2.3.4+5.6.7.(8-4)+........+99.100.101.(102-98)
D=(1.2.3.4+5.6.7.8+.........+99.100.101.102)-(1.2.3.4+5.6.7.8+....+98.99.100.101)
D=98.99.100.101
( 1/5 + x ) mũ 2= 9/16
<=>(1/5+x)mũ 2=(3/4)mũ 2
=>1/5+x=3/4
<=>x=3/4-1/5
<=>x=11/20
vậy x=..............
\(\left(\frac{1}{5}+x\right)^2=\frac{9}{16}\)
<=>\(\left(\frac{1}{5}+x\right)^2-\left(\frac{3}{4}\right)^2=0\)
<=>\(\left(\frac{1}{5}+x-\frac{3}{4}\right)\cdot\left(\frac{1}{5}+x+\frac{3}{4}\right)=0\)
<=>\(\left(x-\frac{11}{20}\right)\cdot\left(x+\frac{19}{20}\right)=0\)
<=>\(\orbr{\begin{cases}x-\frac{11}{20}=0\\x+\frac{19}{20}=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{11}{20}\\x=-\frac{19}{20}\end{cases}}\)
Vậy ...............
\(3\frac{3}{5}x-x=4,5\)
\(\Rightarrow\frac{16}{5}x-\frac{5}{5}x=\frac{9}{2}\)
\(\Rightarrow x\left(\frac{16}{5}-\frac{5}{5}\right)=\frac{9}{2}\)
\(\Rightarrow\frac{11}{5}x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}:\frac{11}{5}\)
\(\Rightarrow x=\frac{45}{22}\)