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Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
\(x^{2018}-x^{18}=0\)
\(x^{18}.\left(x^{2018}-1\right)=0\)
\(=>\orbr{\begin{cases}x^{18}=0\\x^{2018}-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b) 275 > 81x
<=> 315 > 34x
<=> 15 > 4x
<=> x < 15 /4
c) 1252+x > 258
<=> 53(2+x) > 516
<=> 3(2+x) > 16
<=> 6 + 3x > 16
<=> 3x > 10
<=> x > 10/3
d) 5x . 5x+1 . 5x+2 <= 100...0 ( 18 số 0 ) : 218
<=> 5x+x+1+x+2 <= 1018 : 218
<=> 53x+3 <= 518
<=> 3x+3 <= 18
<=> 3x <= 15
<=> x <= 5
( <= là bé hơn hoặc bằng )
a)
x= 43
b) 2X-12=8
2X =8+12
2X=20
X=20:2
x =10
c)45:(3X-17)=32
45 : (3X-17)=9
3X-17=45:9
3X-17=5
3X=5+17
3X=22
x=22:3
x= 7,33
d)(2X-8)x2=24
( 2X-8)x2 =16
2X-8 =16:2
2X-8 =8
2X =8+8
2X =16
x =16:2
X =8
Đúng thì tk nếu sai thì thôi
Làm ẩu ^^
a) 5^x=5^78:5^14(lấy 78-14)
5^x=5^64
=> x=64
b) 7^x.7^2=7^21
7^x=7^21:7^2
7^x=7^19
=> x=19
(4.x.25).64=3.68
4.x.32 = 3.68:64
4.x.32= 3.68-4
4.x.32= 3.64
4.x.32= 3.1396
4.x.32 = 3888
x.32=3888:4
x=972:32
x=30,375
ko bt co dung ko
\(\left(4x.2^5\right).6^4=3.6^8\)
\(\Rightarrow4x.2^5=3.6^8\div6^4\)
\(\Rightarrow4x.32=3.6^4\)
\(\Rightarrow4x.32=3888\)
\(\Rightarrow4x=3888\div32\)
\(\Rightarrow4x=121,5\)
\(\Rightarrow x=121,5\div4\)
\(\Rightarrow x=30,375\)
Vậy \(x=30,375\)
Hok tốt !!!!!!!! ^^
a, \(x^5-x^2=0\)
\(\Rightarrow x^2\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^3=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
b, \(x^{2020}-x^{2019}=0\)
\(\Rightarrow x^{2019}\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2019}=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
c, \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^4-1\right]\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^4-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
a) \(x^5-x^2=0\)
\(\Rightarrow x^2\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^3-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^3=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x\in\left\{0;1\right\}\)
b) \(x^{2020}-x^{2019}=0\)
\(\Rightarrow x^{2019}\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2019}=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x\in\left\{0;1\right\}\)
Câu c tương tự nhé em!
Chúc em học tốt nhé!