\(71+65.4=x+140:x+260\)

\(331=x+140:x+260\)

\(331-260=x+260:x\)

\(71=x+260:x\)

\(71-x=260:x\)
tới đây tui ko biết

b) \(B=x-x^2\)

\(B=-\left(x^2-x\right)\)

\(B=-\left(x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)\)

\(B=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vậy \(Min_B=\frac{1}{4}\) khi \(x-\frac{1}{2}=\Rightarrow x=\frac{1}{2}\)

a) \(A=4x-x^2+3\)
\(A=-\left(x^2-4x-3\right)\)

\(A=-\left(x^2-4x+4-4-3\right)\)

\(A=-\left[\left(x-2\right)^2-7\right]\)

\(A=-\left(x-2\right)^2+7\)

Vậy \(Max_A=7\) khi \(x-2=0\Rightarrow x=2\)

a) \(2^{27}=\left(2^3\right)^9=8^9\)

\(3^{18}=\left(3^2\right)^9=9^9\)

b) ta có : \(2^{27}=\left(2^3\right)^9=8^9và3^{18}=\left(3^2\right)^9=9^9\)

vì : \(8^9< 9^9\)

nên: \(2^{27}< 3^{18}\)

a) \(\sqrt{169}=13\) và \(\sqrt{196}=14\)

bài 3 :
a) \(A=\frac{\sqrt{72}}{\sqrt{2}}+2\frac{\sqrt{27}}{\sqrt{3}}-3\frac{\sqrt{28}}{\sqrt{63}}=\frac{22}{3}\)tương tự

a)\(\frac{16}{2^n}=2\)

\(\Rightarrow2^n=\frac{16}{2}=8=2^3\)

\(\Rightarrow n=3\)

b) \(\frac{\left(-3\right)^n}{81}=27\)

\(\Rightarrow\left(-3\right)^n=-27.81=-2187=\left(-3\right)^7\)

\(\Rightarrow n=7\)
c) thiếu đề

a) \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)

b) \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=-\frac{2560}{3}\)
e) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2=\frac{17}{12}.\left(\frac{1}{20}\right)^2=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

f) \(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3=2:\left(-\frac{1}{6}\right)^3=2:-\frac{1}{216}=-432\)

có