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\(a,\) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+\left(2.3\right)^9.2^3.3.5}{2^{12}.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{\left(2^{12}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(=\frac{\left(2^{12}.3^{10}\right).6}{\left(2^{11}.3^{11}\right).5}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{2.2}{5}\)
\(=\frac{4}{5}\)
\(b,\) \(\frac{2^{15}.9^4}{6^3.8^3}\)
\(=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)
\(=\frac{2^{15}.3^8}{2^{12}.3^3}\)
\(=2^3.3^5\)
\(=8.243\)
\(=1944\)
Chúc bạn học tốt ^^
a) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.120}{\left(2^3\right)^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+6^9.120}{2^{12}.3^{12}-6^{11}}=\frac{6^{10}.4+6^{10}.20}{6^{12}-6^{11}}=\frac{6^{10}.\left(4+20\right)}{6^{11}.\left(6-1\right)}=\frac{6^{11}.4}{6^{11}.5}=\frac{4}{5}\)
b) \(\frac{2^{15}.9^4}{6^3.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=1944\)
c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(4.2\right)^{10}+4^{10}}{\left(2^3\right)^4+4^6.4^5}=\frac{4^{10}.2^{10}+4^{10}}{2^{12}+4^6.4^5}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.2^{10}}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(1+2^{10}\right)}=\frac{4^{10}}{4^6}=4^4=256\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)
\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+2^9.3^9.120}{\left(2^3\right)^4.3^{12}-2^{11}.3^{11}}\)=\(\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}\)=\(\frac{120}{3^2-2^3.3^3}\)=\(\frac{120}{9-8.27}\)=\(\frac{120}{-207}\)=\(\frac{-40}{69}\)
tk cho mk nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{5^{10}.9^{10}.5^{20}}{5^{15}.5^{15}.3^{15}}=\frac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)
b) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=-3^3=-27\)
c) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\frac{2.6}{3.5}=\frac{4}{5}\)
d) \(\frac{3.11+42}{5^3}=\frac{33+42}{5^3}=\frac{75}{5^3}=\frac{5^2.3}{5^3}=\frac{3}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{1.81^2}{243}.\frac{3^2}{3^2}=\frac{6561}{243}.1=27\)
b, \(4^6.256^2.2^4=2^{12}.2^{16}.2^4=2^{32}\)
c) \(A=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)(Mình rút gọn lun cho nhanh nhé ) \(\Rightarrow A=\frac{4}{5}\)
d) \(\Rightarrow B=70\)k cho mình nha Cô Nàng Họ Dương
Đây nhé : ý a,b mình đã giải thích rồi
c) \(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{12}{15}=\frac{4}{5}\)\(\frac{4}{5}\)
d) \(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}=2.5.7=70\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Ta có : \(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^{10}.26}=\frac{78}{26}=3\)
Ta có : \(B=\frac{4^6.9^5.6^9.120}{8^4.3^{12}+6^{11}}=\frac{4^6.9^5.120.6^9}{2^{12}.3^{12}+6^{11}}=\frac{4^6.9^5.6^9.120}{6^{12}+6^{11}}=\frac{6^9.9^5.4^6.120}{6^{11}\left(6+1\right)}\)
\(=\frac{9^5.4^6.120}{6^2.7}=\frac{\left(9.4\right)^5.480}{36.7}=\frac{36^4.480}{7}\)
( câu b đến đây hết rút được gồi / có chi sai thông cảm hen )