a ) 7,5 : x = 2,25 : \(4\frac{1}{6}\)
b ) 49x : 10.5 = \(3\frac{3}{4}\) : \(3\frac{1}{8}\)
c ) 0,06 : x = x : 24
d ) 6 : x = 6,5 : ( -29, 25 )
? Tìm x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)4,5:0,3=2,25:(0,1.x)
2,25:(0,1.x)=15
0,1.x=0,15
x=0,15:0,1
x=1,5
Vậy x=1,5
b)8:\(\left(\frac{1}{4}.x\right)\)=2:0,02
8:\(\left(\frac{1}{4}.x\right)\)=100
\(\left(\frac{1}{4}.x\right)\)=\(\frac{8}{100}\)
x=\(\frac{8}{100}\):\(\frac{1}{4}\)
x=\(\frac{8}{25}\)
Vậy x=\(\frac{8}{25}\)
c)\(3:2\frac{1}{4}=\frac{3}{4}:\left(6.x\right)\)
\(\frac{3}{4}:\left(6.x\right)=\frac{4}{3}\)
6.x=\(\frac{3}{4}:\frac{4}{3}\)
6.x=\(\frac{9}{16}\)
x=\(\frac{9}{16}\):6
x=\(\frac{3}{32}\)
Vậy x=\(\frac{3}{32}\)
a.
\(4,5\div0,3=2,25\div\left(0,1\times x\right)\)
\(2,25\div\left(0,1\times x\right)=15\)
\(0,1\times x=\frac{2,25}{15}\)
\(0,1\times x=0,15\)
\(x=\frac{0,15}{0,1}\)
\(x=1,5\)
b.
\(8\div\left(\frac{1}{4}\times x\right)=2\div0,02\)
\(8\div\left(\frac{1}{4}\times x\right)=100\)
\(\frac{1}{4}\times x=\frac{8}{100}\)
\(\frac{1}{4}\times x=\frac{2}{25}\)
\(x=\frac{2}{25}\div\frac{1}{4}\)
\(x=\frac{2}{25}\times4\)
\(x=\frac{8}{25}\)
c.
\(3\div2^1_4=\frac{3}{4}\div\left(6\times x\right)\)
\(\frac{3}{4}\div\left(6\times x\right)=3\div2,25\)
\(\frac{3}{4}\div\left(6\times x\right)=\frac{4}{3}\)
\(6\times x=\frac{3}{4}\div\frac{4}{3}\)
\(6\times x=\frac{3}{4}\times\frac{3}{4}\)
\(6\times x=\frac{9}{16}\)
\(x=\frac{9}{16}\div6\)
\(x=\frac{9}{16}\times\frac{1}{6}\)
\(x=\frac{3}{32}\)
\(\frac{1}{3}x:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}x=\frac{35}{12}\)
\(\Rightarrow x=\frac{35}{4}\)
a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(=\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}\cdot x=\frac{35}{8}\cdot\frac{2}{3}\)
\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}=\frac{35}{4}\)
b) \(4,5:0,3=2,5:\left(0,1\cdot x\right)\)
\(=15=2,5:\left(0,1\cdot x\right)\)
\(\Rightarrow0,1\cdot x=2,25:15\)
\(\Rightarrow x=\frac{3}{20}:0,1=\frac{3}{2}=1,5\)
c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)
\(8:\left(\frac{1}{4}\cdot x\right)=100\)
\(\Rightarrow\frac{1}{4}\cdot x=8:100\)
\(\Rightarrow x=\frac{2}{25}:\frac{1}{4}=\frac{8}{25}=0,32\)
d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)
\(=\frac{4}{3}=\frac{3}{4}:\left(6:x\right)\)
\(\Rightarrow6\cdot x=\frac{3}{4}:\frac{4}{3}\)
\(\Rightarrow x=\frac{9}{16}:6=\frac{3}{32}=0,09375\)
a. 6,5 -9/4:/x+1/3\=/-2\
6,5-9/4:/x+1/3\=2
9/4:/x+1/3\=6,5-2
9/4:/x+1/3\=4,5
/x+1/3\=9/4:4,5
/x+1/3\=1/2
x+1/3=1/2 hoặc x+1/3= -1/2
x= 1/2-1/3 x= -1/2-1/3
x= 1/6 x= -5/6
Vậy x=1/6 hoặcx= -5/6
b. 2-/3/2x-1/4\ = /-5/4\
2-/3/2x-1/4\=5/4
/3/2x-1/4\=2-5/4
/3/2x-1/4\=3/4
3/2x-1/4=3/4 hoặc 3/2x-1/4= -3/4
3/2x=3/4+1/4 3/2x= -3/4+1/4
3/2x=1 3/2x= -1/2
x=1:3/2 x= -1/2:3/2
x=2/3 x= -1/3
Vậy x=2/3 hoặc x= -1/3
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
làm 1 bài mẫu nha
a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{25}{9}\)
\(\Rightarrow x=\frac{25}{9}:\frac{1}{3}\)
\(\Rightarrow x=\frac{25}{3}\)
các bài sau dễ lắm
a) \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{2}{3}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:\frac{1}{3}\)
\(\Rightarrow x=5\)
Ai học lp 7 rùi giải giúp tôi vs