Bài 1 :

a) \(M=\dfrac{1}{2}x^2y.\left(-4\right)y\)

\(\Rightarrow M=-2x^2y^2\)

Khi \(x=\sqrt[]{2};y=\sqrt[]{3}\)

\(\Rightarrow M=-2.\left(\sqrt[]{2}\right)^2.\left(\sqrt[]{3}\right)^2\)

\(\Rightarrow M=-2.2.3=-12\)

b) \(N=xy.\sqrt[]{5x^2}\)

\(\Rightarrow N=xy.\left|x\right|\sqrt[]{5}\)

\(\Rightarrow\left[{}\begin{matrix}N=xy.x\sqrt[]{5}\left(x\ge0\right)\\N=xy.\left(-x\right)\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}N=x^2y\sqrt[]{5}\left(x\ge0\right)\\N=-x^2y\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)

Khi \(x=-2< 0;y=\sqrt[]{5}\)

\(\Rightarrow N=-x^2y\sqrt[]{5}=-\left(-2\right)^2.\sqrt[]{5}.\sqrt[]{5}=-4.5=-20\)

2:

Tổng của 4 đơn thức là;

\(A=11x^2y^3+\dfrac{10}{7}x^2y^3-\dfrac{3}{7}x^2y^3-12x^2y^3=0\)

=>Khi x=-6 và y=15 thì A=0

Do \(x^2\ge0;\forall x\)

\(\Rightarrow\sqrt{x^2+9}-2025\ge\sqrt{0+9}-2025=-2022\)

C là đáp án đúng

a

1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)

\(A=5x^3-15x+7x^2-5x^3-7x^2\)

\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)

\(A=-15x\)

Thay \(x=-5\) vào A ta được:

\(-15\cdot-5=75\)

Vậy: ....

2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)

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

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

\(B=-4x^3-3x\)

Thay \(x=10,y=-1\) vào B ta được:

\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)

Vậy: ....

help

f: ĐKXĐ: \(\dfrac{2x-1}{2-x}>=0\)

=>\(\dfrac{2x-1}{x-2}< =0\)

=>\(\dfrac{1}{2}< =x< 2\)

g: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>0\end{matrix}\right.\Leftrightarrow3< =x< 5\)

h: ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\x+5>=0\end{matrix}\right.\Leftrightarrow x>=1\)

Bài 2: 

a: (2x-1)(x²+5x-4)

\(=2x^3+10x^2-8x-x^2-5x+4\)

\(=2x^3+9x^2-13x+4\)

b: \(=-\left(10x^2+15x-8x-12\right)\)

\(=-\left(10x^2+7x-12\right)\)

\(=-10x^2-7x+12\)

c: \(=7x^2-28x-\left(14x^3-7x^2+28x+3x^2-3x+12\right)\)

\(=7x^2-28x-14x^3+4x^2-25x-12\)

\(=-14x^3+11x^2-53x-12\)

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)

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

\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)

\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)

\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)

c: \(x^3-x^2+2=0\)

=>\(x^3+x^2-2x^2+2=0\)

=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)

=>x+1=0

=>x=-1

Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)

a) Thay `x=1/2` vào A được:

`A=(5. 1/2 -7)(2. 1/2 +3)-(7 . 1/2 +2)(1/2 -4)=5/4`

b) Thay `x=2;y=-2` vào B được:

`B=(2+2.2)(-2-2.2)+(2-2.2)(-2+2.2)=-40`.

a) Với \(x=\dfrac{1}{2}\) ta được:

\(\Leftrightarrow A=\left(\dfrac{5.1}{2}-7\right)\left(\dfrac{2.1}{2}+3\right)-\left(\dfrac{7.1}{2}+2\right)\left(\dfrac{1}{2}-4\right)\)

\(\Leftrightarrow A=-\dfrac{9}{2}.4-\dfrac{11}{2}.\left(-\dfrac{7}{2}\right)\)

\(\Rightarrow A=\dfrac{5}{4}\)

b) Với \(x = 2; y = - 2 \) ta được :

\(\Leftrightarrow B=\left(2-2\left(-2\right)\right)\left(\left(-2\right)-2.2\right)+\left(2+2\left(-2\right)\right)\left(\left(-2\right)+2.2\right)\)

\(\Leftrightarrow B=-40\)