x mũ 2 (x-1)-16 (x-1)= 0 giúp vs ạ
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6 tháng 8 2022
b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>x=2 hoặc x=1
e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)
=>x(x-5)(x-6)=0
hay \(x\in\left\{0;5;6\right\}\)
6 tháng 8 2022
b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
=>x=1 hoặc x=2
24 tháng 9 2020
1) x3 - 3x2 = 0
<=> x2( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
2) 5x( x - 2020 ) - x + 2020 = 0
<=> 5x( x - 2020 ) - ( x - 2020 ) = 0
<=> ( x - 2020 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-2020=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{1}{5}\end{cases}}\)
3) ( 3x - 5 )2 = ( x + 1 )2
<=> ( 3x - 5 )2 - ( x + 1 )2 = 0
<=> [ ( 3x - 5 ) - ( x + 1 ) ][ ( 3x - 5 ) + ( x + 1 ) ] = 0
<=> ( 3x - 5 - x - 1 )( 3x - 5 + x + 1 ) = 0
<=> ( 2x - 6 )( 4x - 4 ) = 0
<=> \(\orbr{\begin{cases}2x-6=0\\4x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
4) ( x2 - 2x )2 - 2( x - 1 )2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x + 1 ) + 2 = 0
<=> ( x2 - 2x )2 - 2x2 + 4x - 2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x ) = 0
<=> ( x2 - 2x )( x2 - 2x - 2 ) = 0
<=> \(\orbr{\begin{cases}x^2-2x=0\\x^2-2x-2=0\end{cases}}\)
+) x2 - 2x = 0 <=> x( x - 1 ) = 0 <=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) x2 - 2x - 2 = 0
<=> x2 - 2x + 1 - 3 = 0
<=> ( x2 - 2x + 1 ) = 3
<=> ( x - 1 )2 = ( ±√3 )2
<=> \(\orbr{\begin{cases}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\)
24 tháng 4 2019
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
24 tháng 4 2019
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 10 2023
Bài 1:
a) \(4^{x+2}+4^x=68\)
\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)
\(\Rightarrow4^x\cdot17=68\)
\(\Rightarrow4^x=\dfrac{68}{17}\)
\(\Rightarrow4^x=4\)
\(\Rightarrow4^x=4^1\)
\(\Rightarrow x=1\)
b) \(5\cdot2^{x+4}-3\cdot2^x=308\)
\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)
\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)
\(\Rightarrow2^x\cdot77=308\)
\(\Rightarrow2^x=\dfrac{308}{77}\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
c) \(4\cdot3^{x+1}+7\cdot3^x=513\)
\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)
\(\Rightarrow3^x\cdot19=513\)
\(\Rightarrow3^x=\dfrac{513}{19}\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
d) \(5^{x+4}-5^x=3120\)
\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)
\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)
\(\Rightarrow5^x\cdot624=3120\)
\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)
\(\Rightarrow5^x=5\)
\(\Rightarrow5^x=5^1\)
\(\Rightarrow x=1\)
f) \(3\cdot4^{2x+1}-16^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)
\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)
\(\Rightarrow4^{2x}\cdot11=2816\)
\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)
\(\Rightarrow4^{2x}=256\)
\(\Rightarrow\left(2^2\right)^{2x}=2^8\)
\(\Rightarrow2^{4x}=2^8\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Bài 2:
\(2^x+124=5^y\)
\(\Rightarrow5^y-2^x=124\)
\(\Rightarrow5^y-2^x=125-1\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)
Vậy: ....
CN
a) x.(x+1)>0
b) (x-1).(x-2)<0
c) x mũ 3=0
d) (x-1) mũ 5= (x-1) mũ 7
e) x mũ 4 =16
Nhanh ak, giúp mik với
2
12 tháng 1 2017
a)x2>-1
=>x>-1
b)\(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}=>\hept{\begin{cases}x< 1\\x< 2\end{cases}}}\)
c) x3=0
=>x=0
d) chịu
e) x4=16
=>x=2
bài nào mk giải cho bợn đều đúng.....
nhớ tích mk nha
12 tháng 1 2017
E,x^4=16=>x=2
C,x^3=0=>x=0
làm đc 2 câu, hình như bnaj thiếu đk của bài
6 tháng 11 2021
a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
\(x^2\left(x-1\right)-16\left(x-1\right)=0\\ \left(x^2-16\right)\left(x-1\right)=0\\ \left(x-4\right)\left(x+4\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=1\end{matrix}\right.\)
Vậy..........
1+1 bằng mấy ạ