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\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)
\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{23}.3^{23}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{23}.3^{23}}=\frac{6}{2^{11}.3^{13}}=\frac{2.3}{2^{11}.3^{12}}=\frac{1}{2^{10}.3^{11}}=\frac{1}{6^{10}.3}\)
a, 205.510/1005
=205.55.55/1005
=1005.55/1005
=55
=3125
b, (0,9)5/(0,3)6
=(0,3.3)5/0,36
=0,55.35/0,36
=35/0,3
=810
c, 63+3.62+33/-13
=(2.3)3+3.(3.2)2+33/-13
=23.33+3.32.22+33/-13
=33.23+33.22+33/-13
=33(23+22+1)/-13
=27.13/-13
=-27
d, 46.95+69.120/84.312-611
=(22)6.(32)5+(2.3)9.3.23.5/(23)4.312-(2.3)11
=212.310+29.39.3.23.5/212.312-211.311
=212.310+212.310.5/211.311.2.3-211.311
=212.310.(1+5)/211.311(6-1)
=212.310.6/211.311.5
=2.6/3.5
=12/15
=4/5
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)
Tham khảo :
Câu hỏi của ๖ۣۜ๖ۣۜHσàηɠ ๖ۣۜTử ๖ۣۜGĭóッ - Toán lớp 7 - Học trực tuyến OLM
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.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)
\(A=\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+2^9.3^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{-2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{-3.7}=\frac{-4}{7}\)
E = \(\frac{\left(2^2\right)^6.\left(3^2\right) ^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^{13}.3^{10}.5}{-2^{11}.3^{11}.\left(2.3+1\right)}\)
E = \(\frac{2^{12}.3^{10}.\left(1+5\right)}{-2^{11}.3^{11}.7}\)
E = \(\frac{2^{12}.3^{10}.6}{-2^{11}.3^{11}.7}\)
E=\(\frac{-2^{11}.\left(-2\right).3^{10}.6}{-2^{11}.3^{10}.3.7}\)
E = \(\frac{-2.6}{3.7}=-\frac{4}{7}\)
Vậy E = -4/7
Ý F bn lm tương tự nha
thank bn nha