tìm x : (x–2). (x–2)= (–162):(–2)
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3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
a) 162 : 𝑥 + 216 : 𝑥 = 6
(162 + 216) : x = 6
378 : x = 6
x = 378 : 6
x = 63
b) 602 : x + 123 x 2 = 253
602 : x + 246 = 253
602 : x = 253 - 246
602 : x = 7
x = 602 : 7
x = 86
c) 486 : x - 126 :x = 6
(486-126) : x = 6
360 : x = 6
x = 360 : 6
x = 60
d) 725 : x + 175 : x = 5
(725 + 175) : x = 5
900 : x = 5
x = 900 : 5
x = 180
a)\(162:x+216:x=6\)
⇔\(\left(162+216\right):x=6\)
⇔\(378:x=6\)
⇔\(x=63\)
b)\(602:x+123.2=253\)
⇔\(602:x+246=253\)
⇔\(602:x=7\)
⇔\(x=86\)
c)\(486:x-126:x=6\)
⇔\(\left(486-126\right):x=6\)
⇔\(360:x=6\)
⇔\(x=60\)
d)\(725:x+175:x=5\)
⇔\(\left(725+175\right):x=5\)
⇔\(900:x=5\)
⇔\(x=180\)
\(a.162:x+216:x=6\) \(b.602:x+123x2=253\)
\(\left(162+216\right):x=6\) \(602:x+246=253\)
\(378:x=6\) \(602:x=253-246\)
\(x=378:6=63\) \(602:x=7\)
\(x=602:7=86\)
\(c.486:x-126:x=6\) \(d.725:x+175:x=5\)
\(\left(486-126\right):x=6\) \(\left(725+175\right):x=5\)
\(360:x=6\) \(900:x=5\)
\(x=360:6=60\) \(x=900:5=180\)
3x-1+5.3x-1=162
=>3x-1.(1+5)=162
=>3x-1.6=162=>3x-1=162:6=27=33
=>x-1=3=>x=4
5x+5x+2=650
=>5x+5x.52=650
=>5x.(1+25)=650
=>5x.26=650=>5x=650:26=25=52
=>x=2
Vì \(\left(x-5\right)^{2018}\ge0;\left|2y^2-162\right|^{2018}\ge0\Rightarrow\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}\ge0\)
mà \(\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}=0\)
Dấu ''='' xảy ra khi x = 5 ; \(2y^2=162\Leftrightarrow y^2=81\Leftrightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Vì \(\left(x-5\right)^{2018}\ge0\\ \left|2y^2-162\right|^{2018}\ge0\\ \)
Suy ra phương trình dc thỏa mãn khi và chỉ khi x-5 = 0 và 2y^2-162=0
\(\left\{{}\begin{matrix}\left(x-5\right)^{2018}=0\\\left|2y^2-162\right|^{2018}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\2\left(y^2-81\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\pm9\end{matrix}\right.\)
Mk sẽ giải từng câu
\(a)\) \(\left(3x+1\right)\left(x-2\right)>0\)
Trường hợp 1 :
\(\hept{\begin{cases}3x+1>0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}3x>-1\\x>2\end{cases}\Leftrightarrow}\hept{\begin{cases}x>\frac{-1}{3}\\x>2\end{cases}}}\)
\(\Rightarrow\)\(x>2\)
Trường hợp 2 :
\(\hept{\begin{cases}3x+1< 0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}3x< -1\\x< 2\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{-1}{3}\\x< 2\end{cases}}}\)
\(\Rightarrow\)\(x< \frac{-1}{3}\)
Vậy \(x>2\) hoặc \(x< \frac{-1}{3}\) thì \(\left(3x+1\right)\left(x-2\right)>0\)
Chúc bạn học tốt ~
a) (3x+1).(x-2)>0
TH1: 3x+1>0 TH2: x-2>0
3x > -1 x>2
x>-1/3
Vậy x>2
a) 467 + x = 877 – 162
467 + x = 715
x = 715 – 467
x = 248
b) x – 214 = 61 + 345
x – 214 = 406
x = 406 + 214
x = 620
c) x : 5 = 10 × 2
x : 5 = 20
x = 20 × 5
x = 100
d) x × 4 = 30 – 10
x × 4 = 20
x = 20 : 4
x = 5
3^2.6.3^x+1=162
9.6.3^x+1=162
6.3^x+1=162:9
6.3^x+1=18
3^x+1=18:6
3^x+1=3
3^x+1=3^1
Suy ra x=0
vậy x=o thỏa mãn đề bài
(x–2). (x–2)= (–162):(–2)
=> (x-2)2 = 81
=> (x-2)2 = 92
=> x-2 = 9
=> x = 9 + 2
=> x = 11
Vậy x = 11
~k+kb nha~
(x-2)(x-2)=(-162):(-2)
<=> (x-2)2=81
<=> \(\orbr{\begin{cases}\left(x-2\right)^2=9^2\\\left(x-2\right)^2=\left(-9\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=11\\x=-7\end{cases}}}\)
Vaayj x=11; x=-7