5.(22.32).(22)6-2.(2.(22.3)14.34
5.228.318-7.229.318
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\(A=\dfrac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{20}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(=\dfrac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-14\right)}=\dfrac{2.9}{-9}=-2\)
a: =6^2-15*2
=36-30=6
b: =3*25+15*4-3
=72+60
=132
c: =25*2+5
=50+5
=55
d: =5+12=17
e: =7^2(15-5)=49*10=490
f: =4^3*125-125:25
=1000-5
=995
22.3(x+5) - 62 = (23 + 22).22
4.3(x+5) - 36= 48
12(x+5) = 84
x+5 = 84 : 12
x+5 = 7
x = 2
tick mk nha :))
Bài 2
\(a,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)
\(\Rightarrow x\in\left\{4;5;6\right\}\)
\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)
\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)
Mà \(\frac{15}{2}\notin n\)
\(\Rightarrow x\in\left\{7;8\right\}\)
#)Giải :
Bài 1 :
a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)
Bài 2 :
a) \(\left(x-5\right)^2=\left(x-5\right)^6\)
\(\Leftrightarrow x^4-625=x^6-15625\)
\(\Leftrightarrow x^6-x^4=15000\)
\(\Leftrightarrow x^6-x^4=5^6-5^4\)
\(\Leftrightarrow x=5\)
b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow2x-15=1\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
\(\hept{\begin{cases}\\\end{cases}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^{ }^2_{ }\widebat{ }\varnothing\alpha}\)ok
chua