Tìm x
42x - 6 = 1
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a) \(\left(x-6\right)^2=9=3^2\)
\(\Rightarrow x-6=3\) hay \(x-6=-3\)
\(\Rightarrow x=9\) hay \(x=3\)
b) \(4^{2x-6}=1=4^0\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
a) \(\left(x-6\right)^2=9\)
\(\Rightarrow\left(x-6\right)^2=3^2\)
\(\Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=9\end{matrix}\right.\)
b) \(4^{2x-6}=1\)
\(\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
4^2x+1+4^2x=20
4^2x.4^1+4^2x.1=20
4^2x.(4^1+1)=20
4^2x.5=20
4^2x=20:5
4^2x=4
4^2x=2^2
2x=2
x=2:2
x=1
sai đừng trách mk còn đúng thì 1 like
4^2x+1+4^2x=20
4^2x.4^1+4^2x.1=20
4^2x.(4^1+1)=20
4^2x.5=20
4^2x=20:5
4^2x=4
4^2x=2^2
2x=2
a, 42x - 6 = 1
=> 42 x = 7
=> x = 6
b, 5x + 5x + 1 +5x + 2 = 775
=> 15 x + 3 = 775
=> 15 x = 772
=> x = 772/ 15
Câu 1:
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)(1)
Trường hợp 1: x<1
(1) trở thành 1-x+2-x=3
=>3-2x=3
=>x=0(nhận)
Trường hợp 2: 1<=x<2
(1) trở thành x-1+2-x=3
=>1=3(loại)
Trường hợp 3: x>=2
(1) trở thành x-1+x-2=3
=>2x-3=3
=>2x=6
hay x=3(nhận)
=1/x*(1/2+1/6+1/12+1/20+1/30+1/42).
Ta có:
1/2+1/6+1/12+1/20+1/30+1/42.
=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7.
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7.
=1-1/7.
=6/7.
=>1/x*6/7=36.
=>1/x=36:6/7=42.
=>x=1/42.
Vậy x=1/42.
ĐKXĐ: \(\left[{}\begin{matrix}x\ge3+\sqrt{3}\\x\le3-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(3x-7\right)^2}-1=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow\left|3x-7\right|-1=3\sqrt{x^2-6x+6}\)
- Với \(x\ge3+\sqrt{3}\):
\(\Leftrightarrow3x-8=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow9x^2-48x+64=9\left(x^2-6x+6\right)\)
\(\Rightarrow x=-\frac{10}{3}\left(l\right)\)
- Với \(x\le3-\sqrt{3}\)
\(\Leftrightarrow2-x=\sqrt{x^2-6x+6}\)
\(\Leftrightarrow x^2-4x+4=x^2-6x+6\Rightarrow x=1\)
\(ĐKXĐ:\orbr{\begin{cases}x\ge3+\sqrt{3}\\x\le3-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\sqrt{\left(3x-7\right)^2}-1=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow\left|3x-7\right|-1=3\sqrt{x^2-6x+6}\)
- Với \(x\ge3+\sqrt{3}\)
\(\Leftrightarrow3x-8=3\sqrt{x^2-6x+6}\)
\(\Leftrightarrow9x^2-48x+64=9\left(x^2-6x+6\right)\)
\(\Rightarrow x=-\frac{10}{3}\left(l\right)\)
- Với \(x\le3-\sqrt{3}\)
\(\Leftrightarrow2-x=\sqrt{x^2-6x+6}\)
\(\Leftrightarrow x^2-4x+4=x^2-6x+6\)
\(\Rightarrow x=1\) ( t/m)
Chúc bạn học tốt !!!
\(a,\Rightarrow2^{2x-5}=2^5\Rightarrow2x-5=5\Rightarrow x=5\\ b,\Rightarrow4^{2x-1}=4^3\Rightarrow2x-1=3\Rightarrow x=2\\ c,\Rightarrow\left[{}\begin{matrix}2x+2=11\\2x+2=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\\ d,\Rightarrow2^{x^3}=256=2^8\Rightarrow x^3=8\Rightarrow x=2\)
a)
\(2^{2x-5}=2^5\)
2x-5=5
2x=10
x=5
b)
\(4^{2x-1}=4^3\)
2x-1=3
2x=4
x=2
c)
\(\left(2x+2\right)^2=11^2\)
2x+2=11
2x=9
x=9/2
a) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
b) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
c) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2.
=>2x-6=0=>x=3
42x - 6 = 1
42x - 6 = 40
2x - 6 = 0
2x = 6
x = 6 : 2 = 3
vậy ...