Bài 1:

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

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

Bài 2: 

a: \(\Leftrightarrow4x^2-4x+1-4x^2-16x-16=9\)

=>-20x-15=9

=>-20x=24

=>x=-6/5

b: \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

=>9x=18

=>x=2

Bài 1:

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

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

\(A=x^3+y^3\)

Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:

\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)

Bài 2:

a) \(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(A=6x^2+33x-10x-55-6x^2-14x-9x-27\)

\(A=-82\)

b) \(B=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(B=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)

\(B=29\)

c) \(C=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(C=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)\)

\(C=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

\(C=-8\)

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1.

a) \(A=\left(x-1\right)^3-\left(x+4\right)\left(x^2-4x+16\right)+3x\left(x-1\right)\)

\(A=\left(x^3-3x^2+3x-1\right)-\left(x^3+64\right)+\left(3x^2-3x\right)\)

\(A=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)

\(A=\left(x^3-x^3\right)+\left(-3x^2+3x\right)+\left(3x-3x\right)+\left(-1-64\right)\)

\(A=-65\)

Vậy giá trị của biểu thức trên không phụ thuộc vào biến.

b) \(B=\left(x+y-1\right)^3-\left(x+y+1\right)^3+6\left(x+y\right)^2\)

\(B=\left[\left(x+y-1\right)-\left(x+y+1\right)\right].\left[\left(x+y-1\right)^2+\left(x+y-1\right).\left(x+y+1\right)+\left(x+y+1\right)^2\right]+6\left(x+y\right)^2\)

\(B=\left(x+y-1-x-y-1\right).\left[\left(x+y\right)^2-2\left(x+y\right).1+1+\left(x+y\right)^2-1+\left(x+y\right)^2+2\left(x+y\right).1+1\right]+6\left(x+y\right)^2\)

\(B=-2.\left(x^2+2xy+y^2-2x-2y+1+x^2+2xy+y^2-1+x^2+2xy+y^2+2x+2y+1\right)+6\left(x+y\right)^2\)

\(B=-2.\left(3x^2+6xy+3y^2+1\right)+6\left(x+y\right)^2\)

\(B=-2.\left(3x^2+6xy+3y^2\right)-2+6\left(x+y\right)^2\)

\(B=-6\left(x+y\right)^2+6\left(x+y\right)^2-2\)

\(B=-6\left[\left(x+y\right)^2-\left(x+y\right)^2\right]-2\)

\(B=-2\)

Vậy giá trị của biểu thức trên không phụ thuộc vào biến.

2. \(A=x^2+6x+11\)

\(A=x^2+2x.3+3^2+2\)

\(A=\left(x+3\right)^2+2\)

Ta có: \(\left(x+3\right)^2\ge0\)

\(\Rightarrow\left(x+3\right)^2+2\ge2\)

\(\Rightarrow Min_A=2\Leftrightarrow x=-3\)

\(B=4-x^2-x\)

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

\(B=-x^2-x-\dfrac{1}{4}+\dfrac{17}{4}\)

\(B=-\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{17}{4}\)

\(B=-\left(x+\dfrac{1}{2}\right)^2+\dfrac{17}{4}\)

Ta có: \(-\left(x+\dfrac{1}{2}\right)^2\le0\)

\(\Rightarrow-\left(x+\dfrac{1}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\)

\(\Rightarrow Max_B=\dfrac{17}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

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Bài 4 :

Thay x=y+5 , ta có :

a ) ( y+5)*(y5+2)+y*(y-2)-2y*(y+5)+65

=(y+5)*(y+7)+y^2-2y-2y^2-10y+65

=y^2+7y+5y+35-y^2-2y-2y^2-10y+65

= 100

Bài 5 :

A = 15x-23y

B = 2x-3y

Ta có : A-B

= ( 15x -23y)-(2x-3y)

=15x-23y-2x-3y

=13x-26y

=13x*(x-2y) chia hết cho 13 

=> Nếu A chia hết cho 13 thì B chia hết cho 13 và ngược lại 

bài 1: a) \(A=\frac{\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right)}{\frac{a+2}{a-2}}\)

\(A=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{a+2}{a-2}\)

\(A=\left(\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\right)\cdot\frac{a-2}{a+2}\)

\(A=2\cdot\frac{a-2}{a+2}\left(a\ne0;a\ne\pm2\right)\)

b) để A = 1 => \(2\cdot\frac{a-2}{a+2}=1\)

=> 2a - 4 = a + 2

=> a = 6 (thỏa mãn)

bài 2) a) ĐKXĐ: \(x\ne4\)

b) \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(\Leftrightarrow B=\frac{2\sqrt{x}+\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\frac{2\sqrt{x}+4}{x-4}=\frac{2}{\sqrt{x}-2}\)

c) \(B=\frac{2}{\sqrt{3+2\sqrt{3}}-2}\) \(\approx3,69\)

(bạn tự bấm máy tính nhé nhưng theo mình thấy nếu x = 4 + 2\(\sqrt{3}\) hay \(3+2\sqrt{2}\) thì sẽ cho kết quả đẹp hơn, k biết bạn có nhầm đề k nữa!)

bài 3)

a, \(B=\left(\frac{1}{\sqrt{y}+1}-\frac{3\sqrt{y}}{\sqrt{y}-1}+3\right)\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\left(y\ne1;y\ne4\right)\)

\(\Leftrightarrow B=\frac{\sqrt{y}-1-3y-3\sqrt{y}+3y-3}{y-1}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\)

\(\Rightarrow B=\frac{-2\sqrt{y}-4}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\Rightarrow B=\frac{-2}{\sqrt{y}-1}\)

b) y = \(3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

=> B = \(\frac{-2}{\sqrt{\left(1+\sqrt{2}\right)^2}-1}\)

\(\Rightarrow B=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

Bài 6:

a) \(x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)

\(\Leftrightarrow-14x+2=30\)

\(\Leftrightarrow-14x=28\)

\(\Leftrightarrow x=-2\)

d) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow2x=-16\)

\(\Leftrightarrow x=-8\)

Em cần gấp bây h ạ :<

Bài 1:

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

b) \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^2+x^3+x+x^2+1\)

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

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