Đại số lớp 7

\(PT< =>-x^2+10x-25=0\\ < =>-\left(x-5\right)^2=0\)

\(< =>x-5=0\\ < =>x=5\)

\(X=\left(2xy^3\right)^{n-1}\cdot x\cdot\left(x^2y\right)^{n-2}\cdot\left(2xyz\right)^{2n+3}\)

\(=2^{n-1}\cdot x^{n-1}\cdot y^{3n-3}\cdot x\cdot x^{2n-4}\cdot y^{n-2}\cdot2^{2n+3}\cdot x^{2n+3}\cdot y^{2n+3}\cdot z^{2n+3}\)

\(=\left(2^{n-1}\cdot2^{2n+3}\right)\cdot\left(x^{n-1+1+2n-4+2n+3}\right)\cdot\left(y^{3n-3+n-2+2n+3}\right)\cdot z^{2n+3}\)

\(=2^{3n+2}\cdot x^{5n-1}\cdot y^{6n-2}\cdot z^{2n+3}\)

C nha bạn

\(A=-\dfrac{1}{2}xy^3\cdot\left(xyz^2\right)^3+\left(a-2\right)\cdot\left(x^2y^2\right)^2\cdot y^2z^6\)

\(=-\dfrac{1}{2}xy^3\cdot x^3y^3z^6+\left(a-2\right)\cdot x^4y^4\cdot y^2z^6\)

\(=-\dfrac{1}{2}x^4y^6z^6+\left(a-2\right)\cdot x^4y^6z^6\)

\(=x^4y^6z^6\left(-\dfrac{1}{2}+a-2\right)=\left(a-\dfrac{5}{2}\right)\cdot x^4y^6z^6\)

Để A>0 với mọi x,y,z khác 0 thì \(a-\dfrac{5}{2}>0\)

=>\(a>\dfrac{5}{2}\)

\(B=\left(3a-\dfrac{1}{2}\right)xy^4z^2\cdot\left(xy^2\right)^3+\left(a-1\right)\cdot y^2z^2\cdot\left(xy^2\right)^4\)

\(=\left(3a-\dfrac{1}{2}\right)\cdot xy^4z^2\cdot x^3y^6+\left(a-1\right)\cdot y^2z^2\cdot x^4y^8\)

\(=\left(3a-\dfrac{1}{2}\right)\cdot x^4y^{10}z^2+\left(a-1\right)\cdot x^4y^{10}z^2\)

\(=\left(3a-\dfrac{1}{2}+a-1\right)\cdot x^4y^{10}z^2\)

\(=\left(4a-\dfrac{3}{2}\right)\cdot x^4y^{10}z^2\)

Để B<0 thì \(4a-\dfrac{3}{2}< 0\)

=>\(4a< \dfrac{3}{2}\)

=>\(a< \dfrac{3}{2}:4=\dfrac{3}{8}\)

Các cặp đơn thức đồng dạng là: 

\(-2xy^5;6xy^5\)

7;0

=>Có 2 cặp

Đặt \(A=\dfrac{3}{8}xy^3z\cdot\left(-12xy\right)\)

\(=\dfrac{3}{8}\cdot\left(-12\right)\cdot xy^3z\cdot xy\)

\(=-\dfrac{36}{8}x^2y^4z\)

\(=-\dfrac{9}{2}x^2y^4z\)

Thay x=1/2;y=2;z=-3 vào A, ta được:

\(A=-\dfrac{9}{2}\cdot\left(\dfrac{1}{2}\right)^2\cdot2^4\cdot\left(-3\right)\)

\(=\dfrac{27}{2}\cdot\dfrac{1}{4}\cdot16=27\cdot2=54\)

\(\left(-\dfrac{2}{3}x^3y^2\right)\cdot\left(\dfrac{3}{4}xy^3\right)\)

\(=-\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot x^3y^2\cdot xy^3\)

\(=-\dfrac{1}{2}x^4y^5\)

Bậc là 4+5=9