câu a)

Ta có: Góc AOC=góc DOB (đối đỉnh)

Mà góc AOC=góc AOE(tia phân giác)

=>góc DOB=góc AOE(dpcm)

b)Ta có:góc AOE+góc EOF=90 độ

Góc DOB+góc FOD=90 độ

Mà góc AOE=góc DOB

=>góc FOD=góc FOE

Hay OF là phân giác của góc EOD(dpcm)

\(4.\left(-\frac{1}{2}\right)^3-2.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1\)

\(=4.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right).\left(-\frac{1}{2}\right)\right)-2.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right)\right)-3.\left(-\frac{1}{2}\right)\)( sửa ngoặc vuông giúp mk )

\(=4.\left(-\frac{1}{8}\right)-2.\left(\frac{1}{4}\right)-3.\left(-\frac{1}{2}\right)\)

\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)

\(=1+\frac{3}{2}\)

\(=\frac{5}{2}\)

\(4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)^1\)

\(=4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)\)

\(=4\times\left(-\frac{1}{8}\right)-2\times\frac{1}{4}-3\times\left(-\frac{1}{2}\right)\)

\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)

\(=1+\frac{3}{2}\)

\(=\frac{2}{2}+\frac{3}{2}\)

\(=\frac{5}{2}\)

\(4.\left(-\frac{1}{2}\right)^3-3.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1=4.-\frac{1}{8}-3.\frac{1}{4}-3.-\frac{1}{2}\)

                                                                               \(=-\frac{1}{2}-\frac{3}{4}-\left(-\frac{3}{2}\right)\)

                                                                               \(=-\frac{1}{2}-\frac{3}{4}+\frac{3}{2}\)

                                                                               \(=\left(-\frac{1}{2}+\frac{3}{2}\right)-\frac{3}{4}\)

                                                                               \(=1-\frac{3}{4}\)

                                                                               \(=\frac{1}{4}\)

Hok tốt nha^^

x³ + x + 2

\(=x^3+x^2-x^2-x+2x+2\)

\(=x^2\left(x+1\right)-x\left(x+1\right)+2\left(x+1\right)\)

\(\left(x+1\right)\left(x^2-x+2\right)\)

c) x³ + 3²x - 4

\(=x^3-x^2+4x^2-4x+4x-4\)

\(=x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+4x+4\right)\)

\(=\left(x-1\right)\left(x+2^2\right)\)

d)x³y³ + x²y² + 4

\(=x^3y^3-4xy+x^2y^2-4xy+4\)

\(=xy\left(x^2y^2-4\right)+\left(xy+2\right)^2\)

\(=xy\left(xy-2\right)\left(xy+2\right)+\left(xy+2\right)^2\)

\(=\left(xy+2\right)\left(xy\left(xy-2\right)+xy+2\right)\)

\(=\left(xy+2\right)\left(x^2y^2-2xy+xy+2\right)\)

\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)  

e) x³ + 3x²y - 9xy² + 5y³

^{\(=x^3-3x^2y+3xy^2-y^3+6x^2y-12xy^2+6y^3\)}

\(=\left(x-y\right)^3\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^3\left(x-y\right)^2=\left(x-y\right)^2\left(x-y-1\right)\)

Câu hỏi của Nguyễn Khánh Ly - Toán lớp 7 - Học toán với OnlineMath

Em tham khảo link này nhé!

link nào bạn

\(c=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.93}+\frac{1}{93.91}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

\(c=\frac{1}{99.97}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\)

\(c=\frac{1}{99.97}\left(1-\frac{1}{97}\right)\)

\(c=\frac{1}{99.97}.\frac{96}{97}\)

\(c=\frac{32}{33.97.97}\)

bai hoang lan sai roi

A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)

\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)