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1) \(\Rightarrow5x-36=216\)
\(\Rightarrow5x=252\Rightarrow x=\dfrac{252}{5}\)
2) \(\Rightarrow\left[{}\begin{matrix}x-890=0\\74-x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=890\\x=74\end{matrix}\right.\)
(5x-36):18=12
5x-36=216
5x=252
x=50,4
(x-890)(74-x)=0
=> x-890=0 hoặc 74-x=0
x=890 hoặc x=74
a) \(\left(x+74\right)-310=200\)
\(x+74=200+310\)
\(x+74=510\)
\(x=510-74\)
\(x=436\)
b) \(3636:\left(12\cdot x-91\right)=36\)
\(12\cdot x-91=3636:36\)
\(12\cdot x-91=101\)
\(12\cdot x=101+91\)
\(12\cdot x=192\)
\(x=192:12\)
\(x=16\)
Vậy ....
Chú ý rằng |a| = b với b > 0 thì a = b hoặc a = - b.
a ) x ∈ − 5 8 ; 5 8 b ) x ∈ 1 12 ; 1 4
c ) x ∈ − 17 30 ; 9 10 d ) x ∈ − 19 4 ; 11 2
a ) x = 5 12 x = − 5 12 b ) x = 89 14 x = 23 14
c ) x = 69 20 x = − 49 20 d ) x = 22 9 x = − 8 9
a)(-49)-x-13=-6
(-49)-x=-6+13
x=(-49)-7
x=-56
b)74+x+12=5
x=5-12-74
x=-81
a: =>48+338:(6x-5)^2=50
=>338:(6x-5)^2=2
=>(6x-5)^2=169
=>6x-5=13 hoặc 6x-5=-13
=>6x=-8 hoặc 6x=18
=>x=3 hoặc x=-4/3
b: =>120:[16-12(3x-74)]=81-21=60
=>16-12(3x-74)=2
=>12(3x-74)=14
=>3x-74=7/6
=>x=451/18
a: =>338:(6x-5)^2=25*2-48=2
=>(6x-5)^2=169
=>6x-5=13 hoặc 6x-5=-13
=>x=-4/3 hoặc x=3
b: =>120:[16-12(3x-74)]=81-21=60
=>16-12(3x-74)=2
=>12(3x-74)=14
=>3x-74=7/6
=>3x=451/6
=>x=451/18
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left[\frac{x-12}{77}-1\right]+\left[\frac{x-11}{78}-1\right]=\left[\frac{x-74}{15}-1\right]-\left[\frac{x-73}{16}-1\right]\)
\(\Leftrightarrow\frac{x-12-77}{77}+\frac{x-11-78}{78}=\frac{x-74-15}{15}+\frac{x-73-16}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}=0\)
\(\Leftrightarrow\left[x-89\right]\cdot\left[\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right]=0\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89
\(0,48x12+0,24+0,12x74\)
=\(0,12x4x12+0,12x2+0,12x74\)
=\(0,12x\left(48+2+74\right)\)
=\(0,12x124\)
=14,88