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a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(1999^{2x-6}=1\)
\(\Rightarrow1999^{2x-1}=1999^0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
c) \(x^{2002}=x\)
\(\Rightarrow x^{2002}-x=0\)
\(\Rightarrow x.\left(x^{2001}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)
+) \(x=0\)
+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d) \(\left(x-1\right)^2=9\)
\(\Rightarrow x-1=\pm3\)
+) \(x-1=3\Rightarrow x=4\)
+) \(x-1=-3\Rightarrow x=-2\)
Vậy \(x\in\left\{4;-2\right\}\)
e) \(\left(2x-3\right)^2=81\)
\(\Rightarrow2x-3=\pm9\)
+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)
+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)
Vậy \(x\in\left\{6;-3\right\}\)
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