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Ta có:
\(P=2a^{2n+1}-3a^{2n}+5a^{2n+1}-7a^{2n}+3a^{2n+1}\)
\(P=\left(2a^{2n+1}+5a^{2n+1}+3a^{2n+1}\right)+\left(-3a^{2n}-7a^{2n}\right)\)
Suy ra: \(P=10a^{2n+1}+\left(-10a\right)^{2n}\)
Mà \(2n⋮2\)còn \(2n+1\)ko chia hết cho 2
Do đó: \(a>0\)thì P>0
a) ta có: \(M=\left(\frac{1}{3}a-\frac{1}{3}b\right)-\left(a+2b\right)\)
\(M=\frac{1}{3}a-\frac{1}{3}b-a-2b\)
\(M=(\frac{1}{3}a-a)+\left(\frac{-1}{3}b-2b\right)\)
\(M=\frac{-2}{3}a+\frac{-7}{3}b\)
\(N=\frac{1}{3}a-\frac{1}{3}b-\left(a-b\right)\)
\(N=\frac{1}{3}a-\frac{1}{3}b-a+b\)
\(N=\left(\frac{1}{3}a-a\right)+\left(b-\frac{1}{3}b\right)\)
\(N=\frac{-2}{3}a+\frac{2}{3}b\)
\(\Rightarrow M+N=\left(\frac{-2}{3}a+\frac{-7}{3}b\right)+\left(\frac{-2}{3}a+\frac{2}{3}b\right)\)
\(=\frac{-2}{3}a+\frac{-7}{3}b+\frac{-2}{3}a+\frac{2}{3}b\)
\(=\left(\frac{-2}{3}a-\frac{2}{3}a\right)+\left(\frac{-7}{3}b+\frac{2}{3}b\right)\)
\(=\frac{-4}{3}a+\frac{-5}{3}b\)
\(\Rightarrow M+N=\frac{-4}{3}a-\frac{5}{3}b\)
ta có: \(M-N=\left(\frac{-2}{3}a+\frac{-7}{3}b\right)-\left(\frac{-2}{3}a+\frac{2}{3}b\right)\)
\(=\frac{-2}{3}a+\frac{-7}{3}b+\frac{2}{3}a-\frac{2}{3}b\)
\(=\left(\frac{-2}{3}a+\frac{2}{3}a\right)+\left(\frac{-7}{3}b-\frac{2}{3}b\right)\)
\(=0+\frac{-10}{3}b=\frac{-10}{3}b\)
\(\Rightarrow M-N=\frac{-10}{3}b\)
b) ta có: \(M=2a^2+ab-b^2-\left(-a^2+b^2-ab\right)\)
\(M=2a^2+ab-b^2+a^2-b^2+ab\)
\(M=\left(2a^2+a^2\right)+\left(ab+ab\right)+\left(-b^2-b^2\right)\)
\(M=3a^2+2ab+\left(-2b^2\right)\)
\(N=3a^2+b^2-\left(ab-a^2\right)\)
\(N=3a^2+b^2-ab+a^2\)
\(N=\left(3a^2+a^2\right)+b^2-ab\)
\(N=4a^2+b^2-ab\)
rồi bn tính như mk phần a nha!
c) ta có: \(M=\left(x+cy-z\right)+y+x-\left(z-x-y\right)\)
\(M=x+cy-z+y+x-z+x+y\)
\(M=\left(x+x+x\right)+\left(y+y\right)+\left(-z-z\right)+cy\)
\(M=3x+2y+\left(-2z\right)+cy\)
\(N=x-\left(x-\left(y-z\right)-x\right)\)
\(N=x-\left(x-y+z-x\right)\)
\(N=x-x+y-z+x\)
\(N=\left(x-x+x\right)+y-z\)
\(N=x+y-z\)
bn tính giúp mk cộng trừ 2 đa thức M; N luôn nha! mk chỉ rút gọn cho bn thôi
CHÚC BN HỌC TỐT!!!!
\(\left(-\dfrac{1}{3}m^2\right)\left(-24n\right)\left(4mn\right)\)
\(=\left(8nm^2\right)\left(4mn\right)\)
\(=32n^2m^3\)
\(\left(5a\right)\left(a^2b^2\right)\left(-2b\right)\left(-3a\right)\)
\(=\left(5a^35ab^2\right)\left(6ab\right)\)
\(=30a^4b.30a^2b^3\)
Thông cảm mk làm cái này hay sai lắm
a) \(A\left(x\right)=2\left(x^3\right)^n-7\left(x^n\right)^3+8x^{3n-2}.x^2-4x^3x^{3n-3}\)
\(A\left(x\right)=2x^{3+n}-7x^{3+n}+8x^{3n-2+2}-4x^{3+3n-3}\)
\(A\left(x\right)=2x^{3+n}-7x^{3+n}+8x^{3n}-4x^{3n}\)
\(A\left(x\right)=-5x^{3+n}+4^{3n}\)
b) Thay \(x=\frac{-1}{2};n=1\)vào A(x)
\(A\left(\frac{-1}{2}\right)=-5.\left(\frac{-1}{3}\right)^{3+1}+4^{3.1}\)
\(A\left(\frac{-1}{2}\right)=\left(\frac{5}{3}\right)^4+4^3\)
\(A\left(\frac{-1}{2}\right)=\left(\frac{125}{27}\right)+64\)
\(A\left(\frac{-1}{2}\right)=\frac{1934}{27}\)
Mình nhầm ở phần kết quả cuối cùng của câu a
Đáng lẽ phải là \(A\left(x\right)=-6x^{3-n}-4x^{3n}\)
Bạn tính lại phần b giúp mình nhé, sr
`Answer:`
\(-\frac{2}{3}x^3y^2.\left(-\frac{4}{3}a^2bxy\right)^2\)
\(=-\frac{2}{3}x^3y^2.\frac{16}{9}a^4b^2x^2y^2\)
\(=\left(-\frac{2}{3}.\frac{16}{9}\right).a^4.b^2.\left(x^3.x^2\right).\left(y^2.y^2\right)\)
\(=-\frac{32}{27}a^4b^2x^5y^4\)
\(\left(-\frac{1}{2}a^2b^2x^2y\right)^3\left(-2a^3b^{-3}xy^2\right)^2\)
\(=\left(-1a^6b^6x^6y^3\right)\left(4a^6\frac{1}{b^6}x^2y^4\right)\)
\(=\left(-\frac{1}{8}.4\right).\left(a^6.a^6\right).\left(b^6.\frac{1}{b^6}\right)\left(x^6.x^2\right).\left(y^3.y^4\right)\)
\(=-\frac{1}{2}a^{12}x^8y^7\)
\(\left(-\frac{a}{2}\right)^3\left(-2b\right)^2\left(5x^2y^3\right)^2\left(-\frac{1}{5}a^2b^3zy\right)^3\)
\(=\left(-\frac{a^3}{8}\right).4b^2.25x^4.y^6.\left(-\frac{1}{125}a^6b^9z^3y^3\right)\)
\(=\left(-\frac{1}{8}\right).a^3.4b^2.25x^4.y^6.\left(-\frac{1}{125}a^6b^9z^3y^3\right)\)
\(=\left(-\frac{1}{8}.4.25.-\frac{1}{125}\right).\left(a^3.a^6\right).\left(b^2.b^9\right).x^4.\left(y^6.y^3\right).z^3\)
\(=0,1.a^9.b^{11}.x^4.y^9.z^3\)
a. P = 0
b . a = 0