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1. \(\left(-\frac{3}{4}\right)^{3x-2}=\frac{81}{256}\)
\(\left(\frac{-3}{4}\right)^{3x-2}=\left(\frac{-3}{4}\right)^4\)
=> 3x - 2 = 4
=> 3x = 6
=> x = 2
2. \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
=> 3x - 1 = 5
=> 3x = 6
=> x = 2
mk giải ở dưới được 2 câu rùi nhưng ko chắc cau 2 ~~~
5465765756876
Ta có
1,\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)\)
\(\left(x+1\right)\left(3x-1\right)\)
2, \(x^3+2x^2+4x^2+8x+3x+6\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+3\right)\)
\(=\left(x+2\right)\left(x^2+x+3x+3\right)\)
\(=\left(x+2\right)\text{[}x\left(x+1\right)+3\left(x+1\right)\text{]}\)
\(=\left(x+2\right)\left(x+1\right)\left(x+3\right)\)
3,\(x^4+2x^2-3=x^4-x^2+3x^2-3\)
\(=x^2\left(x^2-1\right)+3\left(x^2-1\right)\)
\(\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4,\(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
1/
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\frac{A}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(A-\frac{A}{3}=\frac{2A}{3}=\frac{1}{3}-\frac{1}{3^{100}}\Rightarrow2A=1-\frac{1}{3^{99}}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{99}}<\frac{1}{2}\)
2/ Làm tương tự bài 1
\(x^2=\frac{1}{16}=\left(\frac{1}{4}\right)^1=\left(-\frac{1}{4}\right)^2\)
Vậy có 2 ngiệm x
TH1: \(x=\frac{1}{4}\)
TH2: \(x=-\frac{1}{4}\)
x2=1/16
=>x=1/4; x=-1/4
x5=(2/3)5
=>x=2/3
x4=(3/2)4
=>x=3/2; x=-3/2
1) 5x+5x+2=650
5x.1+5x.52=650
5x(1+52)=650
5x(1+25)=650
5x.26=650
5x=650:26=25=52
=> x=2
2) 3x-1+5.3x-1=162
3x-1.1+5.3x-1=162
3x-1(1+5)=162
3x-1.6=162
3x-1=162:6=27=33
=> x-1=3
x=3+1=4
3) 23x+2=4x+5
23x.22=4x.45
23x+2=(2.2)x.(2.2)5=2x.2x.25.25=22x.25+5=22x.210=22x+10
=> 3x+2=2x+10
3x+2-2x=10
3x-2x=10-2=8
x=8
Tick nha