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bài 2 nhá
a, 7 chia hết cho 2x-1
=> 2x-1 thuộc ư(7) = ( 1,-1,7,-7)
ta có bảng
2x-1/1/-1/7/-7.x/2/0/6/-4
b x-6 chia hết cho x-1
=>x-1-5 chia hết x-1
=>5 chia hết x-1 hay x-1 thuộc ư (5) = ( liệt kê)
tạo bảng tương tự như câu a
tích cho mik nha
46.95 + 69.120 / 84.312 - 611
= 212.315 + 212.310.5 / 212.312 - 211.311
= 212. 310.( 35 + 5 ) / 211.311( 2.3 - 1 )
= 2 . 248 / 3 . 5 = 496/15
\(a,\) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+\left(2.3\right)^9.2^3.3.5}{2^{12}.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{\left(2^{12}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(=\frac{\left(2^{12}.3^{10}\right).6}{\left(2^{11}.3^{11}\right).5}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{2.2}{5}\)
\(=\frac{4}{5}\)
\(b,\) \(\frac{2^{15}.9^4}{6^3.8^3}\)
\(=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)
\(=\frac{2^{15}.3^8}{2^{12}.3^3}\)
\(=2^3.3^5\)
\(=8.243\)
\(=1944\)
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) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(4.2\right)^{10}+4^{10}}{\left(2^3\right)^4+4^6.4^5}=\frac{4^{10}.2^{10}+4^{10}}{2^{12}+4^6.4^5}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.2^{10}}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(1+2^{10}\right)}=\frac{4^{10}}{4^6}=4^4=256\)
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) \(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}=2.5.7=70\)
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)
Bạn 12345678901 xuống lớp 1 học đạo đức làm người nhé bạn. Lịch sự tí đi
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
a. \(\frac{20^5.5^{10}}{100^5}\)= \(\frac{20^5.5^{10}}{20^5.5^5}\)= \(5^5\)=\(3125\)
b. \(\frac{0,9^5}{0,3^6}\)= \(\frac{0,9^5}{0,3^5.0,3}\)= \(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)= \(243.\frac{1}{0,3}\)= \(810\)
c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)