tìm x
c)[3(x+\(\dfrac{1}{2}\))+5(0,6-x)].50%=-x:6\(\dfrac{1}{5}\)
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a: \(\Leftrightarrow x^2=900\)
=>x=30 hoặc x=-30
b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)
=>0,1x=2/3:50/3=2/3x3/50=1/25
=>1/10x=1/25
hay x=1/25:1/10=10/25=2/5
d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)
=>x=12/5 hoặc x=-12/5
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
`(50% . x+5 1/4) . (-2/3) = 2 5/6`
`(1/2 x + 21/4) . (-2/3) = 17/6`
`1/2 x + 21/4 = -17/4`
`1/2 x = -19/2`
`x=-19`
\(^{^{tham}}\) \(^{^{khảo}}\)
\(\left(50\%.+5\dfrac{1}{4}\right).\left(-\dfrac{2}{3}\right)=2\dfrac{5}{6}\)
\(\left(\dfrac{1}{2}x+\dfrac{21}{2}\right).\left(-\dfrac{2}{3}\right)=\dfrac{17}{6}\)
\(\dfrac{1}{2}x+\dfrac{21}{4}=-\dfrac{17}{4}\)
\(\dfrac{1}{2}x=-\dfrac{19}{2}\)
\(x=-19\)
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
a) \(-0,6x-\dfrac{7}{3}=5,4\Leftrightarrow-\dfrac{3}{5}x=5,4+\dfrac{7}{3}\Leftrightarrow x=\dfrac{116}{15}.\left(-\dfrac{5}{3}\right)=-\dfrac{116}{9}\).
b) \(2,8:\left(\dfrac{1}{5}-3x\right)=1\dfrac{2}{5}\Leftrightarrow\dfrac{1}{5}-3x=2,8:\dfrac{7}{5}\Leftrightarrow-3x=2-\dfrac{1}{5}\Leftrightarrow x=\dfrac{9}{5}:\left(-3\right)=-\dfrac{3}{5}\).
Lời giải:
$\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}$
$\frac{x}{3}=\frac{1}{4}-\frac{5}{6}=\frac{-7}{12}$
$x=\frac{-7}{12}.3=\frac{-7}{4}$
-------------------
$\frac{2x}{3}=\frac{6}{x}$ ($x\neq 0$)
$\Rightarrow 2x^2=18$
$x^2=18:2=9=(-3)^2=3^2$
$\Rightarrow x=\pm 3$
---------------------
$\frac{x-1}{14}=\frac{4}{x}$ ($x\neq 0$)
$\Rightarrow x(x-1)=14.4=56$
$x^2-x-56=0$
$(x+7)(x-8)=0$
$\Rightarrow x+7=0$ hoặc $x-8=0$
$\Leftrightarrow x=-7$ hoặc $x=8$
---------------------------
$\frac{-x}{8}=\frac{-50}{x}$ ($x\neq 0$)
$\Rightarrow -x^2=8(-50)$
$x^2=400=20^2=(-20)^2$
$\Rightarrow x=\pm 20$
a, \(\dfrac{3}{4}x=-\dfrac{9}{8}\)
x= \(-\dfrac{3}{2}\)
b, |x| + 0,25= 5,25
|x | = 5
=> x\(\in\){ +- 5}
Ko chắc đúng, kiểm tra trc khi làm
\(\dfrac{2x-1}{3}=\dfrac{-5}{0.6}\)
\(\Leftrightarrow2x-1=-25\)
hay x=-12
a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
\(\Rightarrow\)\(\left[3x+\dfrac{3}{2}+3-5x\right].\dfrac{1}{2}=-x:\dfrac{31}{5}\)
\(\Rightarrow\)\(\left[-2x+\dfrac{9}{2}\right].\dfrac{1}{2}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(\dfrac{-2x}{2}+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(-x+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5}{31}+x\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5+31x}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x\left(-5+31\right)}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x.26}{31}\)
\(\Rightarrow\dfrac{9}{4}.31=x.26\)\(\Rightarrow\dfrac{279}{4}=x.26\)
\(\Rightarrow x=\dfrac{279}{4}:26=\dfrac{279}{4}.\dfrac{1}{26}\)
\(\Rightarrow x=\dfrac{279}{104}\)
Vậy x=\(\dfrac{279}{104}\)