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Lí do = 0 nè:
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)
a,x/8=2/x
=>x2=16
=>x=-4;4
vậy x=-4;4
b,\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
\(\Rightarrow\left(\frac{x-39}{19}+1\right)+\left(\frac{x-49}{29}+1\right)=\left(\frac{x-59}{39}+1\right)+\left(\frac{x-69}{49}+1\right)\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\Rightarrow\left(x-20\right)\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
vì \(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\Rightarrow x-20=0\Rightarrow x=20\)
vậy x=20
\(\frac{x}{8}=\frac{2}{x}\Rightarrow x^2=8.2\)
\(x^2=16\Rightarrow x^2=\left(\pm4\right)^2\Rightarrow x=4;x=-4\)
Học tốt
a. 60%x + 0,4x + x : 3 = 2
0.6x + 0,4x + x : 3 = 2
x(0,6 + 0,4 : 3 ) = 2
\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)
câu B tự làm nha .
= 25/49 *( 21/29-7/29) +24/49 * 15/29
=25/49*14/29+ 24/49*15/29
BẠN TỰ LÀM TIẾP NHA
study well
\(\frac{25}{49}.\frac{21}{29}-\frac{25}{49}.\frac{7}{29}+\frac{24}{49}.\frac{15}{29}\)
\(=\frac{25}{49}.\left(\frac{21}{29}-\frac{7}{29}\right)+\frac{24}{49}.\frac{15}{29}\)
\(=\frac{25}{49}.\frac{14}{29}+\frac{24}{49}.\frac{15}{29}\)
\(=\left(\frac{25}{49}+\frac{24}{49}\right).\left(\frac{14}{29}+\frac{15}{29}\right)\)
\(=1.1=1\)
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
Tìm x biết:
\(\frac{x}{3}-\frac{3}{4}=\frac{1}{12}\)
\(\frac{x}{3}=\frac{1}{12}+\frac{3}{4}\)
\(\frac{x}{3}=\frac{5}{6}\)
\(x=\frac{5}{6}.3\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{199}{230}\)
\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
Vậy \(x=\frac{3}{10}\)
Bài 2: tính
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{5}-\frac{1}{11}\)
\(=\frac{6}{55}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Bài 2:
1/30+1/42+1/56+1/72+1/90+1/110
=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11
=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1/5-1/11=6/55
b)1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50
=49/50
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Công mỗi vế cho 2 ta được:
\(\frac{x-39}{19}+\frac{x-49}{29}+2=\frac{x-59}{39}+\frac{x-69}{49}+2\)
\(\frac{x-39}{19}+1+\frac{x-49}{29}+1=\frac{x-59}{39}+1+\frac{x-69}{49}+1\)
\(\frac{x-39}{19}+\frac{19}{19}+\frac{x-49}{29}+\frac{29}{29}=\frac{x-59}{39}+\frac{39}{39}+\frac{x-69}{49}+\frac{49}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\left(x-20\right).\frac{1}{19}+\left(x-20\right).\frac{1}{29}-\left(x-20\right).\frac{1}{39}-\left(x-20\right).\frac{1}{49}=0\)
Đặt thừa số chung (x-20) ra ngoài
\(\left(x-20\right).\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
\(\text{Vì }\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\text{ nên }x-20=0\Rightarrow x=20\)