sorry em mới học lớp 6 thui

phan van nam ngon thi lam coi

A=\(\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x(2x-5y)}{ 8x(x-3y)} =\frac{2x-5y}{x-3y} \)

\(\frac{x}{y}=\frac{10}{3}<=>10y=3x <=>y=\frac{3}{10}x \)

=>A=(\(2x-\frac{3}{2}x):(x-\frac{9}{10}x) \)

=\(\frac{1}{2}x:\frac{1}{10}x=\frac{1}{2}x.\frac{10}{x}=5 \)

a)\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\)Giá trị của biểu thức trên tại \(a=4;b=-5;c=6\) là:

\(\dfrac{4-5-6}{4-\left(-5\right)+6}=-\dfrac{7}{15}\)

b: \(=\dfrac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\dfrac{2x-5y}{x-3y}\)

Đặt x/10=y/3=k

=>x=10k; y=3k

\(A=\dfrac{2\cdot10k-5\cdot3k}{10k-3\cdot3k}=\dfrac{5k}{k}=5\)

c: \(C=\left(\dfrac{x^3-y^3-x^3-y^3}{\left(x+y\right)\left(x-y\right)}\right):\dfrac{x^2-y^2-x^2}{x+y}\)

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

\(=\dfrac{20}{9-10}=-20\)

\(\frac{x}{y}=10\Rightarrow x=10y\)

\(M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}\)

\(=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}\)

yx​=10⇒x=10y

M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}M=8x2−24xy16x2−40xy​=8x(x−3y)8x(2x−5y)​=x−3y2x−5y​

=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}=10y−3y2.10y−5y​=715​
 

Câu 2

\(\frac{x}{y}\)nha bạn

a) Để giá trị của biểu thức B xác định

\(\Leftrightarrow\hept{\begin{cases}x^2-7x\ne0\\x^2+7x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-7\right)\ne0\\x\left(x+7\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne\pm7\end{cases}}}\)

Vậy ................

b) \(B=\left(\frac{7x+1}{x^2-7x}+\frac{7x-1}{x^2+7x}\right):\frac{x^2+1}{x^2-49}\)

\(B=\left(\frac{7x+1}{x\left(x-7\right)}+\frac{7x-1}{x\left(x+7\right)}\right).\frac{\left(x+7\right)\left(x-7\right)}{x^2+1}\)

\(B=\frac{\left(7x+1\right)\left(x+7\right)+\left(7x-1\right)\left(x-7\right)}{x\left(x-7\right)\left(x+7\right)}.\frac{\left(x-7\right)\left(x+7\right)}{x^2+1}\)

\(B=\frac{7x^2+49x+x+7+7x^2-49x-x+7}{x\left(x^2+1\right)}\)

\(B=\frac{14x^2+14}{x\left(x^2+1\right)}\)

\(B=\frac{14}{x}\)

c) Để \(B\inℤ\)

\(\Leftrightarrow\frac{14}{x}\inℤ\)

\(\Leftrightarrow x\inƯ\left(14\right)\)

\(Ư\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

Vậy ...................

a, Ta có : \(\frac{3y}{4}=\frac{3y}{4}.1=\frac{3y}{4}.\frac{2x}{2x}=\frac{6xy}{8x}\) ( đpcm )

b, Ta có : \(6x^2y=6x^2y\)

=> \(3x^2.2y=\left(-3x^2\right).\left(-2y\right)\)

=> \(\frac{-3x^2}{2y}=\frac{3x^2}{-2y}\) ( đpcm )

c, Ta có : \(6x-6y=6x-6y\)

=> \(6x-6y=-6y+6x\)

=> \(6\left(x-y\right)=-6\left(y-x\right)\)

=> \(2\left(x-y\right).3=-2\left(y-x\right).3\)

=> \(\frac{2\left(x-y\right)}{3\left(y-x\right)}=\frac{-2}{3}\) ( đpcm )