a) pt a <=> 3x+1=0     hoặc        x-2000=0         hoặc      3x+6000=0

           <=> x=-1/3      hoặc        x=2000            hoặc      x=-2000

a: 128-2x=-8-x

=>-2x+x=-8-128

=>-x=-136

=>x=136

b: \(x-\left(-15\right)\cdot2=55-7\cdot\left(-3\right)\)

=>x+30=55+21

=>x+30=76

=>x=76-30

=>x=46

c: \(66-\left\lbrack5-\left(-x\right)\right\rbrack=169\)

=>66-[5+x]=169

=>x+5=66-169=-103

=>x=-103-5

=>x=-108

d: (8-2x)(8+x)=0

=>2(4-x)(8+x)=0

=>(x-4)(x+8)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-8\end{array}\right.\)

e: \(7-\left\lbrack x^2+\left(-9\right)\right\rbrack=-20\)

=>\(x^2-9=7+20=27\)

=>\(x^2=36\)

=>\(\left[\begin{array}{l}x=6\\ x=-6\end{array}\right.\)

gì thế

CAU HOI CHU GI NUA

Số số hạng:

[ (2n - 1) - 1] : 2 + 1 = n (số)

⇒\(\frac{2n.n}{2}=169\)

=> 2n² = 338

=> n² = 338:2

=> n² = 169

=> n² = 13²

=> n = 13

a)\(\left(2x-1\right)^5=32\)

\(\Rightarrow\left(2x-1\right)^5=2^5\)

\(\Rightarrow2x-1=2\)

\(\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)

b)\(\left(2x+1\right)^2=169\)

\(\Rightarrow\left(2x+1\right)^2=13^2=\left(-13\right)^2\)

\(\Rightarrow2x+1=13\) hoặc \(2x+1=-13\)

\(\Rightarrow2x=12\) hoặc \(2x=-14\)

\(\Rightarrow x=6\) hoặc \(x=-7\)

c)\(x^{100}=x\)

\(\Rightarrow x^{100}-x=0\)

\(\Rightarrow x\left(x^{99}-1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^{99}-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^{99}=1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)

a, (2x-1)^5=32

(2x-1)^5=2^5

2x-1=2

2x=2+1

2x=3

x=3:2

x=1,5

Vậy x=1,5

b, (2x+1)^2=169

(2x+1)^2=13^2

2x+1=13

2x=13-1

2x=12

x=12:2

x=6

Vậy x=6

c, x^100=x

=>x=0

x=1

ko đưa câu hỏi linh tinh lên diễn đàn

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