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\(\left(3-4x\right)^2=25=5^2\)
\(\Rightarrow3-4x=5\)
\(\Rightarrow4x=3-5=-2\Rightarrow x=-\frac{1}{2}\)
\(\left(2x-\frac{1}{4}\right)^2=16=4^2\)
\(\Rightarrow2x-\frac{1}{4}=4\Rightarrow2x=4+\frac{1}{4}=\frac{17}{4}\)
\(\Rightarrow x=\frac{17}{4}:2=\frac{17}{4}.\frac{1}{2}=\frac{17}{8}\)
Đề số 3 bị sai.
\(\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=-\frac{5}{2}\)
(3-4x)2=25
3-4x=5
4x=3-5
4x=-2
x=-2:4
x=-0,5
b)(2x-1/42)=16
2x-1/4=4
2x=4+1/4
2x=4,25
x=2,125
c) cái này x ở đâu vậy bn
d) (2x+5)2=0
2x+5=0
2x=0+5
2x=5
x=5:2
x=5/2
Nhớ k cho mk nha
a) \(4^{x+3}-248=2^{x+1}\)
\(\Leftrightarrow2^{2x+6}-248=2^{x+1}\)
\(\Leftrightarrow2^{2x+6}-2^{x+1}=248\)
\(\Leftrightarrow2^{x+1}\left(2^{x+5}-1\right)=248=2^3.31=2^2.62=2.124=1.248\)
Thay vào nha
1.
\(\left(2x+3^{ }\right)^4=16\)
⇔\(\left(2x+3\right)^4=2^4\)
⇔\(2x+3=2\)
⇔\(2x=5\)
⇔\(x=\frac{5}{2}\)
Vậy....
a) \(2^{4-2x}=2^{16}\)(biến đổi \(16^4\Rightarrow2^{16}\))
=> 4 - 2x = 16
2x = 4 - 16
2x = -12
x = -12 : 2
x = -6
\(\left[\frac{x}{\left(x+4\right)\left(x-4\right)}-\frac{x-4}{x\left(x+4\right)}\right]:\frac{2\left(x-2\right)}{x\left(x+4\right)}\)\(=\left[\frac{x^2-\left(x-4\right)^2}{x\left(x+4\right)\left(x-4\right)}\right].\left[\frac{x\left(x+4\right)}{2\left(x-2\right)}\right]\)\(=\left(\frac{x^2-x^2+8x-16}{x\left(x+4\right)\left(X-4\right)}\right).\frac{x\left(x+4\right)}{2\left(x-2\right)}=\frac{8\left(x-2\right).x\left(x+4\right)}{x\left(x+4\right)\left(x-4\right).2\left(x-2\right)}=\frac{4}{x-4}\)
\(\Rightarrow2^{4-2x}=2^{16}\\ \Rightarrow4-2x=16\\ \Rightarrow2x=-12\\ \Rightarrow x=-6\)
\(2^{4-2x}=16^4\)
\(\Rightarrow2^{4-2x}=\left(2^4\right)^4\)
\(\Rightarrow2^{4-2x}=2^{16}\)
\(\Rightarrow4-2x=16\)
\(\Rightarrow-2x=12\)
\(\Rightarrow x=-6\)
Vậy: \(x=-6\)