Thay x = 1 và y = - 1 vào biểu thức A sau khi rút gọn ta có:

\(=-\left(-1\right)^5\cdot1+\left(-1\right)^2\cdot1+\left(-1\right)^5\cdot1=1+1-1=1\)

=−(−1)5⋅1+(−1)2⋅1+(−1)5⋅1=1+1−1=1

= 3/4x⁵y

\(\left(\dfrac{1}{2}-\dfrac{3}{4}+1\right)x^5y=\dfrac{3}{4}x^5y\)

Chọn D

1.

\(\frac{-2}{3}x^3y^4.\left(\frac{-5}{9}x^5y\right).3y^7=\left[\left(\frac{-2}{3}\right).\left(\frac{-5}{9}\right).3\right]\left(x^3y^4x^5yy^7\right)=\frac{10}{9}x^8y^{12}\ge0\)

Vậy 3 đơn thuc trên không thể có cùng gt âm (vì nếu cùng âm thì tích của chúng phải âm)

2.

\(A+B=7x^2-5xy+2y^7+\left(5x^2+3xy-y\right)\)

\(=7x^2-5xy+2y^7+5x^2+3xy-y\)

\(=\left(7x^2+5x^2\right)+\left(-5xy+3xy\right)+2y^7-y\)

\(=12x^2-2xy+2y^7-y\)

A-B tương tự

Chọn A

Ta có: -6x5y + 7x5y + (-3x5y) + x5y = -x5y.

Để x5y chia hết cho 2 và 5 thì y = 0

Để x50 chia hết cho 3 thì x = 1;4;7

Vậy {x;y} = {1;4;7;0}

a. x⁴ : xⁿ = x^{4 - n}

b. xⁿ : x⁵ = x^{n - 5}

c. \(\left(3x^4y^3+\dfrac{1}{2}x^3y^2+x^5y\right):4x^ny^n\)

= \(3x^4y^3:4x^ny^n+\dfrac{1}{2}x^3y^2:4x^ny^n+x^5y:4x^ny^n\)

= \(\dfrac{3}{4}x^{4-n}y^{3-n}+\dfrac{1}{8}x^{3-n}y^{2-n}+\dfrac{1}{4}x^{5-n}y^{1-n}\)

a: \(x^4:x^n=x^{4-n}\)

b: \(x^n:x^5=x^{n-5}\)

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

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

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)