tìm x biết :\(\dfrac{6}{x}=\dfrac{24}{x-27}\)
\(\dfrac{-2}{9}\)và\(\dfrac{6}{-27}\) b:\(\dfrac{-1}{-5}\)và\(\dfrac{4}{25}\)
Các cặp phân số sau có bằng nhau ko?vì sao?
Bài3: Tìm số nguyên X biết
a)\(\dfrac{-28}{35}\)=\(\dfrac{16}{x}\)
b)\(\dfrac{x+7}{15}\)=\(\dfrac{-24}{36}\)
giúp mình với ae cứu tôi ae cứu tôi :((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
Bài 2:
a: -2*(-27)=54
6*9=54
=>Hai phân số này bằng nhau
b: -1/-5=1/5=5/25<>4/25
Bài 3:
a: =>16/x=-4/5
=>x=-20
b: =>(x+7)/15=-2/3
=>x+7=-10
=>x=-17
a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)
\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)
Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)
b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)
\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)
Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)
Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)
Bài 3
a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(x=\dfrac{35.16}{-28}\)
\(x=-20\)
b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\left(x+7\right).36=15.\left(-24\right)\)
\(36x+252=-360\)
\(36x=-360-252\)
\(36x=-612\)
\(x=\dfrac{-612}{36}\)
\(x=-17\)
Tìm x biết: a) \(\dfrac{6}{-x}=\dfrac{x}{-24}\) b) \(x-\dfrac{7}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
c)\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\) d) \(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
e) \(\dfrac{9}{x}=\dfrac{-35}{105}\) f) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
Tìm x,y biết :
6) 3x=4y và 2x + 3y = 7
7) \(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}\) và x-y+z=36
8) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\) và 3x-2y+2z = 24
7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36
Nên theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6
\(\Rightarrow\)x=6.5=30
y=6.6=36
z=6.7=42
vậy x=30,y=36,z=42
Tìm x biết: a) x + \(\dfrac{2}{3}=\dfrac{4}{27}\) b) \(\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{1}{4}x+\dfrac{1}{6}\)
c) \(\dfrac{13}{10}x-\dfrac{5}{2}=\dfrac{7}{2}\) d) (3\(x\) + 2) \(\left(\dfrac{-2}{5}x-7\right)=0\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a) x = 4/27 - 2/3
x = -14/27
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
Bài 1:
a)\(\left|x-2\right|\)+\(\left|1-\dfrac{x}{2}\right|\)=0 b)\(\left(x-\dfrac{1}{3}\right)^3\)=\(\dfrac{-8}{27}\)
c)\(\dfrac{x^2}{6}\)=\(\dfrac{24}{25}\) c)\(\dfrac{x-1}{x+5}\)=\(\dfrac{6}{7}\)
Giúp mik làm bài này với ạ mik đang cần gấp tối nay mik phải nộp rồi mong mn giúp đỡ mik. Mik cảm ơn mn
tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)
c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
BT3: Tìm x, biết
18) \(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
\(\Rightarrow\left(\dfrac{x+24}{1996}+1\right)+\left(\dfrac{x+25}{1995}+1\right)+\left(\dfrac{x+26}{1994}+1\right)+\left(\dfrac{x+27}{1993}+1\right)+\left(\dfrac{x+2036}{4}-4\right)=0\)\(\Rightarrow\dfrac{x+2020}{1996}+\dfrac{x+2020}{1995}+\dfrac{x+2020}{1994}+\dfrac{x+2020}{1993}+\dfrac{x+2020}{4}=0\)\(\Rightarrow\left(x+2020\right)\left(\dfrac{1}{9996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)
\(\Rightarrow x+2020=0\Rightarrow x=-2020\)
Tìm x
a)\(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
b)\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
=> 6(2x-7) = 9(x-3)
=> 12x - 42 = 9x - 27
=> 12x - 9x = -27 + 42
=> 3x = 15
=> x = 5
Vậy x = 5
b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
=> -7(x + 27) = 6(x + 1)
=> -7x - 189 = 6x + 6
=> -7x - 6x = 6 + 189
=> -13x = 195
=> x = -15
Vậy x = -15
a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)
\(\Leftrightarrow12x-42=9x-27\)
\(\Leftrightarrow12x-9x=-27+42\)
\(\Leftrightarrow3x=15\)
hay x=5
Vậy: x=5
b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x+189\)
\(\Leftrightarrow6x+7x=189-6\)
\(\Leftrightarrow13x=183\)
hay \(x=\dfrac{183}{13}\)
Vậy: \(x=\dfrac{183}{13}\)
Bài 4. Tìm x biết:
a. \(\dfrac{x}{5}=\dfrac{2}{5},\dfrac{3}{8}=\dfrac{6}{x},\dfrac{1}{9}=\dfrac{x}{27}\)
b. \(\dfrac{4}{x}=\dfrac{8}{6},\dfrac{3}{x-5}=\dfrac{-4}{x+2},\dfrac{x}{-2}=\dfrac{-8}{x}\)
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3
b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x2 => 42 = x2 => x=4
Tick cho mình đi