Tìm số hữu tỉ x trong các tỉ lệ thức :
a ) 0,4 : x = x : 0,9
b )\(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
c ) \(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6x+7\right)\)
d ) \(\frac{37x}{x+13}=\frac{3}{7}\)
Tìm số hữu tỉ x trong tỉ lệ thức :
a ) 0,4 ; x = x : 0,9
b ) \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
c ) \(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6x+7\right)\)
d ) \(\frac{37-x}{x+13}=\frac{3}{7}\)
TÌm số hữu tỉ x trong tỉ lệ thức:
a) 0,4 : x = x : 0,9
b) \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
c) \(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6x+7\right)\)
d) \(\frac{37-x}{x+13}=\frac{3}{7}\)
a, 0,4 : x = x : 0,9
<=> x2 = 0,4 . 0,9
<=> x2 = 0,36
<=> x = 0,6 hoặc -0,6
b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow10=26\div\left(2x-1\right)\)
\(\Leftrightarrow2x-1=\frac{13}{5}\)
\(\Leftrightarrow2x=\frac{18}{5}\)
\(\Leftrightarrow x=\frac{9}{5}\)
c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow6x+7=4\)
\(\Leftrightarrow6x=-3\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d, \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow-10x=-220\)
\(\Leftrightarrow x=22\)
Tìm số hữu tỉ x trong tỉ lệ thức sau
a) 0,4:x=x:0,9 b)\(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
c)\(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6x+7\right)\) d) \(\frac{37-x}{x+13}=\frac{3}{7}\)
Cho tỉ lệ thức \(\frac{3x-y}{x+y}=\frac{3}{4}\). Tính giá trị của tỉ số \(\frac{X}{y}\)
Tìm số hữu tỉ x trong tỉ lệ thức :
a) \(0.2:1\frac{1}{5}=\frac{2}{3}:\left(6.x+7\right)\) b) \(\frac{37-x}{x+13}=\frac{3}{7}\)
Giúp mik với ~!!!!!!!!~
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
tìm x biết :
a , 0,4 : x = x : 0,9
b, \(b,13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
c, \(c,\frac{37-x}{x+13}=\frac{3}{7}\)
a) 0,4:x=x:0,9
0,4.0,9=x2
0,36 =x2
\(\sqrt{0,36}\)=x
0,6 =x
a) 0,4 : x = x : 0,9 x^2=0,9 . 0,4 x^2=(0,6)^2 =>x=+_6
tìm x\(\in\) Q,biết:
a,\(\frac{x}{2}\left(\frac{3x}{5}\right)-\frac{13}{5}=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
b,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
c,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
d,\(\frac{13}{x-1}+\frac{5}{2x+2}-\frac{6}{3x-3}\)
e,\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Ai bt câu nào làm câu đấy nhé ! Làm hết cũng đc nhé!
Bài 1 : Tìm x biết
a) \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)
b) \(0,2:1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)
c) \(\frac{37-x}{x+13}=\frac{3}{7}\)
d) \(\frac{x-1}{x+5}=\frac{6}{7}\)
e) \(2\frac{2}{\frac{3}{0,002}=}\frac{1\frac{1}{9}}{x}\)
Bài 2 : Tìm x,y,z biết:
a) \(\frac{x}{7}=\frac{4}{13}\)và x + y = 40
b) 3x = 2y , 7y = 57 và x - y + z = 32
c) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{7-4}{4}\)và 2x + 3y - z = 50
Bài 3 .
a) 6,88 : x =12:27
b) \(8\frac{1}{3}\div11\frac{2}{3}=13:\left(2x\right)\)
Giải giúp mk
Mk đng cần gấp
1.
a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)
<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26
<=> 10 + 26 = 13x
<=> 13x = 36
<=> x = \(\frac{36}{13}\)
b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)
<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)
<=> x = \(\frac{1}{7}\)
c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)
<=> (37 - x) . 7 = 3.(x + 13)
<=> 119 - 7x = 3x + 39
<=> -7x - 3x = 39 - 119
<=> -10x = -80
<=> x = 8
d) \(\frac{x-1}{x+5}=\frac{6}{7}\)
<=> 7(x - 1) = 6(x + 5)
<=> 7x - 7 = 6x + 30
<=> 7x - 6x = 30 + 7
<=> x = 37
e)
2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)
<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)
<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)
Bài 2. đề sai
Bài 3.
a) 6,88 : x = \(\frac{12}{27}\)
<=> x = 6,88 : \(\frac{12}{27}\)
<=> x = 15,48
b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x
<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x
<=> \(\frac{5}{7}=13:2x\)
<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)
<=> x = 9,1
a)\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
b)\(\left(\frac{-2}{3}x-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
c)\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
d)\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
e)\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}2\right)\)
k)\(\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
m)\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
n)\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Giải giúp mk
mk đang cần ghấp trong hôm nay
cám ơn mọi người
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)