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3.x + 2.x + x = 300
x.(3+2+1) = 300
x.6 = 300
x=300 : 6
x = 50
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
6.x-5=613
6.x=613+5
6.x=618
x=618:6
x=103
12.(x-1)=0
x-1=12.0
x-1=0
x=1+0
x=1
0:x=0
x=0
a,6x-5=613
6x=618
x=103
b,12(x-1)=0
x-1=0
x=1
c,0:x=0
với TH này mọi giá trị của x thỏa mãn
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{7}\cdot\frac{7}{10}\cdot\cdot\cdot\frac{97}{100}\)
\(=\frac{1.4.7.10...97}{4.7.10.13...100}\)
\(=\frac{1}{100}\)
4/3 * [ 7/2 - 1 1/4 ] - 3/4 * x = 2
4/3 * [ 7/2 - 5/4 ] - 3/4 * x = 2
4/3 * [ 14/4 - 5/4] - 3/4 * x = 2
4/3 * 9/4 - 3/4 * x = 2
3 - 3/4 * x = 2
3/4 * x = 3 - 2
3/4 * x = 1
x = 1 : 3/4
x = 4/4 x 4/3 = 4/3
=> x = 4/3
5 * x - 4 * x = 24 * 5
5 * x - 4 * x = 120
x * ( 5 - 4 ) = 120
x * 1 = 120
x = 120 : 1
x = 120
Ủng hộ tui nha! ^^
a) 12 : x + 5 = 33
12 : x = 33 - 5
12 : x = 28
x = 12 : 28
x = 1/4
Vậy x = 1/4
b) 5*(x + 19) = 170 - 50
5*(x+19) = 120
x + 19 = 120 : 5
x + 19 = 24
x = 24 - 19
x = 5
Vậy x = 5
c) 11*(x - 6) - 11 = 4*x
11*(x-7) = 4x
11x - 77 = 4x
11x - 4x = 77
7x = 77
x = 11
Vậy x = 11
a, \(2x+3=15\Leftrightarrow x=6\)
b, \(2\left(42-x\right)=23\Leftrightarrow84-2x=23\Leftrightarrow x=\frac{61}{2}\)
c, \(2,5x+2,3x=144\Leftrightarrow x\left(2,5+2,3\right)=144\Leftrightarrow x=30\)
2(x-1)=6(10-x)
<=>2x-2=60-6x
<=>2x+6x=60+2
<=>8x=62
<=>x=62:8
<=>x=7,75
=> 2x - 2 = 60 - 6x
=> 2x + 6x = 2 + 60
=> 8x = 62
=> x = 7,75
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
a) 0,99 * x + x / 100 = 4,75
0,99 * x + x * 0,01 = 4,75
x * ( 0,99 + 0,01 ) = 4,75
x * 1 = 4,75
x = 4,75