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a: =>3[(2x-1)^2-4]=49*125:175+196=231
=>(2x-1)^2-4=77
=>(2x-1)^2=81
=>2x-1=9 hoặc 2x-1=-9
=>x=5 hoặc x=-4
b: \(\Leftrightarrow2\cdot3^x\cdot3-4^3=7^2\cdot\left(27-25\right)\)
=>\(6\cdot3^x=49\cdot2+64=162\)
=>3^x=27
=>x=3
Lời giải:
a.
$3[(2x-1)^2-4]-14^2=7^2.5^3:175=35$
$3[(2x-1)^2-4]=35+14^2=231$
$(2x-1)^2-4=231:3=77$
$(2x-1)^2=77+4=81=9^2=(-9)^2$
$\Rightarrow 2x-1=9$ hoặc $2x-1=-9$
$\Rightarrow x=5$ hoặc $x=-4$
b.
$2.3^{x+1}-4^{10}:4^7=(7^5:7^3).(3^3-5^2)=7^2.2=98$
$2.3^{x+1}-4^3=98$
$2.3^{x+1}=98+4^3=162$
$3^{x+1}=162:2=81=3^4$
$\Rightarrow x+1=4$
$\Rightarrow x=3$
-1/2.3x+3/4=-2.25
-1/2.3x+3/4=9/4
-1/2.3x =9/4-3/4
-1/2.3x =3/2
3x =3/2:-1/2
3x =-3
x =-3:3
x =-1
vậy x=-1
\(-\frac{1}{2}.3x+\frac{3}{4}=-2,25\)
\(\Rightarrow\frac{1}{2}.(-3).x+\frac{1}{2}.\frac{3}{2}=\frac{9}{4}\)
\(\Rightarrow\frac{1}{2}.\left(-3x+\frac{3}{2}\right)=\frac{9}{4}\)
\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{4}:\frac{1}{2}\)
\(\Rightarrow-3x+\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-1\)
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
a) \(\frac{-1}{2}.3x+\frac{3}{4}=-2,25\)
\(\frac{-1}{2}.3x+\frac{3}{4}=\frac{9}{4}\)
\(\frac{-1}{2}.3x=\frac{9}{4}-\frac{3}{4}\)
\(\frac{-1}{2}.3x=\frac{6}{4}=\frac{3}{2}\)
\(3x=\frac{3}{2}:\frac{-1}{2}\)
\(3x=-3\)
\(x=\left(-3\right):3\)
\(x=-1\)
b) \(13-3\left|x-2\right|=10\)
\(3\left|x-2\right|=13-10\)
\(3\left|x-2\right|=3\)
\(\left|x-2\right|=3:3\)
\(\left|x-2\right|=1\)
\(\Rightarrow\hept{\begin{cases}\left|x-2\right|=1\\\left|x-2\right|=-1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=1\end{cases}}}\)