Tìm x: 3.2^x+2^x=16
(3x-2)^2+1/4=1/2
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KT
1
LT
24 tháng 7 2018
a,=>2^x.4=16=>2^x=4=>x=2
b,=>(3x-2)^2=1/4=>3x-2=1/2=>3x=5/2=>x=5/6
22 tháng 9 2021
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
2 tháng 7 2023
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
TP
1
11 tháng 11 2018
\(2^{x+2}-3.2^x=16\)
=> \(2^x.2^2-3.2^x=16\)
=> \(2^x.\left(2^2-3\right)=16\)
=> \(2^x.1=2^4\)
=> x = 4
\(\left(\frac{1}{5}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
=> \(\left(\frac{1}{5}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
=> \(\orbr{\begin{cases}\frac{1}{5}-\frac{3}{2}x=\frac{3}{2}\\\frac{1}{5}-\frac{3}{2}x=-\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x=\frac{1}{5}-\frac{3}{2}\\\frac{3}{2}x=\frac{1}{5}+\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x=-\frac{13}{10}\\\frac{3}{2}x=\frac{17}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{-13}{15}\\x=\frac{17}{15}\end{cases}}\)
26 tháng 9 2015
a) \(-0,6^0+\frac{1}{2}.2-3x=-\frac{1}{4}\)
\(\Leftrightarrow-1+1-3x=-\frac{1}{4}\Leftrightarrow-3x=-\frac{1}{4}\Leftrightarrow3x=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}:3=\frac{1}{12}\)
b)\(2^{x-2}+22=3.2^x\Leftrightarrow3.2^x-2^{x-2}=22\Leftrightarrow2^{x-2}\left(3.2^2-1\right)=22\)
\(\Leftrightarrow2^{x-2}.11=22\Leftrightarrow2^{x-2}=2\Leftrightarrow x-2=1\Leftrightarrow x=3\)
c) \(\left(x-1\right)^2=\sqrt{\left(-\frac{9}{16}\right)^2}\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\Leftrightarrow\left(x-1\right)^2=\left(\frac{3}{4}\right)^2\)
TH1: x - 1 = 3/4 => x = 3/4 + 1 => x = 7/4
Th2: x - 1 = - 3/4 => x = -3/4 +1 => x = 1/4
d) \(\Leftrightarrow\sqrt{x^2+2}=12-5=7\Leftrightarrow x^2+2=7^2\Leftrightarrow x^2=49-2\Leftrightarrow x^2=47\)
\(x=\sqrt{47};x=-\sqrt{47}\)
2 tháng 7 2021
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1