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\(80-7\left(x-7\right)=95-3^2.6\)
\(\Rightarrow80-7\left(x-7\right)=95-54\)
\(\Rightarrow80-7\left(x-7\right)=41\)
\(\Rightarrow7\left(x-7\right)=80-41\)
\(\Rightarrow7\left(x-7\right)=39\)
\(\Rightarrow x-7=\dfrac{39}{7}\)
\(\Rightarrow x=\dfrac{39}{7}+7\)
\(\Rightarrow x=\dfrac{39}{7}+\dfrac{49}{7}\)
\(\Rightarrow x=\dfrac{88}{7}\)
\(80-7\left(x-7\right)=95-3^2.6\)
\(80-7\left(x-7\right)=95-9.6\)
\(80-7\left(x-7\right)=95-54\)
\(80-7\left(x-7\right)=41\)
\(7\left(x-7\right)=80-41\)
\(7\left(x-7\right)=49\)
\(x-7=49:7\)
\(x-7=7\)
\(x=7+7\)
\(\Rightarrow x=14\)
\(\text{a,}4+8+12+...+80=\frac{\left(80+4\right).20}{2}=840\)
\(b,80-76+72-68+...+8-4=\left(80+72+64+...+8\right)-\left(76+68+60+4\right)\)
\(=\frac{\left(80+8\right).10}{2}-\frac{\left(76+4\right).10}{2}\)
\(40\)
Câu c tương tự câu b
\(\frac{x+5}{95}+\frac{x+10}{90}+\frac{x+15}{85}+\frac{x+20}{80}=-4\)
<=> \(\frac{x+5}{95}+1+\frac{x+10}{90}+1+\frac{x+15}{85}+1+\frac{x+20}{80}+1=0\)
<=> \(\frac{x+100}{95}+\frac{x+100}{90}+\frac{x+100}{85}+\frac{x+100}{80}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{90}+\frac{1}{85}+\frac{1}{80}\right)=0\)
<=> \(x+100=0\) (do 1/95 + 1/90 + 1/85 + 1/80 khác 0)
<=> \(x=-100\)
Vậy...
\(2^{10}:64\cdot16\)
\(=2^{10}:2^6\cdot2^4\)
\(=2^{10-6+4}\)
\(=2^8\)
\(2^{10}.64.16\\ =2^{10}.2^6.2^4\\ =2^{10+6+4}=2^{20}\)
d)37.7+43.7+80.3
=(37+43).7+80.3
=80.7+80.3
=80.(7+3)
=80.10=800
95 : 80 =95
nha bạn
= 9^5 nha bn