Tổng diện tích các tấm bìa của Bắc:

(43+1) : 2 x 22 = 484(cm²)

Tổng diện tích các tấm bìa của Bình:

(46+2):2 x 23= 552(cm²)

Tổng diện tích các tấm bìa của Bình lớn hơn tổng diện tích các tấm bìa của Bắc là:

552 - 484 = 68(cm²)

Đ.số: 68cm²

a: Sửa đề: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

ĐKXĐ: x∉{0;2;-2;3}

Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(=\left\lbrack\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right\rbrack:\frac{x\left(x-3\right)}{x^2\cdot\left(2-x\right)}\)

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

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{-x\left(x-2\right)}{x-3}\)

\(=\frac{-4x^2-8x}{x+2}\cdot\frac{-x}{x-3}=\frac{-4x\left(x+2\right)}{x+2}\cdot\frac{-x}{x-3}=\frac{4x^2}{x-3}\)

b: Để A>0 thì \(\frac{4x^2}{x-3}>0\)

=>x-3>0

=>x>3

c: |x-7|=4

=>\(\left[\begin{array}{l}x-7=4\\ x-7=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=11\left(nhận\right)\\ x=3\left(loại\right)\end{array}\right.\)

Thay x=11 vào A, ta được:

\(A=\frac{4\cdot11^2}{11-3}=\frac{4\cdot121}{8}=\frac{121}{2}\)

\(a,ĐKXĐ:\\ \left[{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\\ b,P=0\\ \Leftrightarrow\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=0\\ \Leftrightarrow\dfrac{3x\left(x+1\right)}{3\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x}{x-2}=0\\ \Leftrightarrow x=0\left(TM\right)\)

Vậy tại X=0 thì P=0

a) Để P xác định thì: \(\left[{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

b) \(P=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)

Để \(P=0\) thì: \(\dfrac{3x}{2x-6}=0\)

\(\Leftrightarrow3x=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

ĐKXĐ: \(x\ne0;x\ne\pm1\)

\(\dfrac{3}{2x^2+2x}+\dfrac{2x-1}{x^2-1}-\dfrac{2}{x}\)

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

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

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

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

\(=\dfrac{1}{2x\left(x-1\right)}\)

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

Kết bạn với tôi đi ,tôi cô đơn quá 

Lời giải:
b. 

$(3x+2)(3x-2)-9x(1+x)=9x^2-4-(9x+9x^2)=9x^2-4-9x-9x^2=-4-9x$

c.

$\frac{1}{x-2}-\frac{1}{x+2}+\frac{2x}{x^2-4}$

$=\frac{x+2}{(x-2)(x+2)}-\frac{x-2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}$

$=\frac{x+2-(x-2)+2x}{(x-2)(x+2)}=\frac{4+2x}{(x-2)(x+2)}$

$=\frac{2(x+2)}{(x-2)(x+2)}=\frac{2}{x-2}$

a: \(\frac{y}{4}-2+\frac{15}{4y}\)

\(=\frac{y-8}{4}+\frac{15}{4y}=\frac{y^2-8y+15}{4y}=\frac{\left(y-5\right)\left(y-3\right)}{4y}\)

\(\frac{y}{2}+\frac{6}{y}-\frac72=\frac{y-7}{2}+\frac{6}{y}=\frac{y\left(y-7\right)+12}{2y}\)

\(=\frac{y^2-7y+12}{2y}=\frac{\left(y-3\right)\left(y-4\right)}{2y}\)

Ta có: \(M=\frac{\frac{y}{4}-2+\frac{15}{4y}}{\frac{y}{2}+\frac{6}{y}-\frac72}\)

\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}:\frac{\left(y-3\right)\left(y-4\right)}{2y}\)

\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}\cdot\frac{2y}{\left(y-3\right)\left(y-4\right)}=\frac{y-5}{2\cdot\left(y-4\right)}\)

b: \(N=\frac{3b-\frac{1}{9b^2}}{1+\frac{1}{3b}+\frac{1}{9b^2}}\)

\(=\frac{27b^3-1}{9b^2}:\frac{9b^2+3b+1}{9b^2}\)

\(=\frac{\left(3b-1\right)\left(9b^2+3b+1\right)}{9b^2}\cdot\frac{9b^2}{9b^2+3b+1}\)

=3b-1

\(199^3-199\\=199(199^2-1)\\=199(199-1)(199+1)\\=199\cdot198\cdot200\)

Vì \(199\cdot198\cdot200⋮200\) nên \(199^3-199⋮200\)

Bắn tung tóe

\(x^2-3\\ =x^2-\left(\sqrt{3}\right)^2\\ =\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

$x^2-3$

$=x^2-(\sqrt3)^2$

$=(x-\sqrt3)(x+\sqrt3)$

ΔAHB vuông tại H

=>\(AH^2+HB^2=AB^2\)

=>\(AH^2=6^2-3,6^2=\left(6-3,6\right)\left(6+3,6\right)=2,4\cdot9,6=4,8^2\)

=>AH=4,8(cm)

Xét ΔAHB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\)

=>\(AE=4,8^2:6=3,84\left(\operatorname{cm}\right)\)

Xét tứ giác AEHF có \(\hat{AEH}=\hat{AFH}=\hat{FAE}=90^0\)

nên AEHF là hình chữ nhật

=>HF=AE

=>HF=3,84(cm)

\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\left(dkxd:x,y\ne0\right)\)

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

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{x^3+y^3}{x^2y^2}\)