Tìm x biết
x mũ 2 phần 3=9 phần x
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8 tháng 9 2021
\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
NT
0
PT
1
6 tháng 11 2021
1) \(\Rightarrow x^2\left(x^{2004}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{2004}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
2) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
TH
1
YN
2 tháng 3 2022
`Answer:`
a, `4x^2-24x+36=(x-3)^3`
`<=>4(x^2-6x+9)-(x-3)^3=0`
`<=>4(x-3)^2-(x-3)^3=0`
`<=>(x-3)^2.(4-x+3)=0`
`<=>(x-3)^2.(7-x)=0`
`<=>x-3=0` hoặc `7-x=0`
`<=>x=3` hoặc `x=7`
b, `(8x^3-7x^2):x^2=3x+\sqrt{\frac{9}{25}}`
`<=>8x^3:x^2-7x^2:x^2=3x+\sqrt{\frac{9}{25}}`
`<=>8x-7=3x+\sqrt{\frac{9}{25}}`
`<=>8x-7=3x+3/5`
`<=>8x=3x+\frac{38}{5}`
`<=>8x-3x=3x+\frac{38}{5}-3x`
`<=>5x=\frac{38}{5}`
`<=>x=\frac{38}{25}`
XO
23 tháng 7 2021
a) \(x^2-\frac{1}{49}=0\)
<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)
Vậy x = \(\pm\frac{1}{7}\)
b) \(64-\frac{1}{4}x^2=0\)
<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)
<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)
Vậy \(x=\pm16\)
c) 9x2 + 12x + 4 = 0
<=> (3x + 2)2 = 0
<=> 3x + 2 = 0
<=> x = -2/3
Vậy x = -2/3
e) \(x^2+\frac{1}{4}=x\)
<=> \(x^2-x+\frac{1}{4}=0\)
<=> \(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
23 tháng 7 2021
d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)
Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng
\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
16 tháng 9 2021
\(x^{20}=x\\ \Leftrightarrow x^{20}-x=0\\ \Leftrightarrow x\left(x^{19}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 7 2023
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
x mũ 2/3=9/x
=>x mũ 2.x=9.3
=>x mũ 3=27
=>27=x mũ 3
=>27=3.3.3
=>27=3 mũ 3
=>x mũ 3=3 mũ 3
=>x=3
Vậy x=3