Tìm x biết :
a) \(\left[x-3\right]=5\)
b) \(\left(x+2\right)^2=81\)
c) \(5^x+5^{x+2}=650\)
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\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
a) 2x - 17 = 35 : 32
2x - 17 = 35 - 2
2x - 17 = 33
= 27
2x = 27 +17
2x = 44
x = 44 :2
vậy x = 22
b, ( 19 - x ) .2 - 20 = 23
= 8
( 19 - x ) .2 = 8 + 20
( 19 - x ) .2 = 28
19 - x = 28 :2
19 - x = 14
x = 19 - 14
vậy x = 5
a)2x-17=243:9
2x-17=27
2x=27+17
2x=44
vậy x=44
b)(19-x).2-20=8
(19-x).2=20+8=28
19-x=28:2
19-x=14
x=19-14
x=5
vậy x=5
c)Bạn tách 3^2 thành 3^1.3^2 nhá . Tương tự tách các phần rồi rút gọn là xong .mik nhắn trên máy tính nó khó ko nhanh được ý ạ bạn thông ca,r
CHUC BAN HOC TOT
a/ \(\left(x+2\right)^2=81\)
\(\Rightarrow\left(x+2\right)^2=9^2;\left(x+2\right)^2=\left(-9\right)^2\)
+) \(x+2=9\)
\(\Rightarrow x=7\)
+) \(x+2=-9\)
\(\Rightarrow x=-11\)
b/ \(5^2+5^{x+2}=650\)
\(\Rightarrow5^x+5^x.5^2=650\)
\(\Rightarrow5^x\left(25+1\right)=650\)
\(\Rightarrow5^x=\frac{650}{26}=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
a) \(\left(x+2\right)^2=81\)
\(\Rightarrow\left(x+2\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow x+2=\pm9\)
\(\Rightarrow\left[\begin{matrix}x+2=-9\\x+2=9\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=-9-2\\x=9-2\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=-11\\x=7\end{matrix}\right.\)
Vậy \(x=-11\) hoặc \(x=7\) thì thỏa mãn đề bài.
b) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x+5^x\times5^2=650\)
\(\Rightarrow5^x\times\left(1+5^2\right)=650\)
\(\Rightarrow5^x\times\left(1+25\right)=650\)
\(\Rightarrow5^x\times26=650\)
\(\Rightarrow5^x=650\div26\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\) thì thỏa mãn đề bài.
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
x-3=5
=>x=8
b) (x+2)2= 81
=> x+2 thuộc { 9;-9}
=> x thuộc {7; -11}
c) 5x+ 5x+2= 650
=> 5x.(1+25)=650
=>5x= 25
=> x=2
vậy x=2
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