Cho x-2y=6
a) Tính K= \(\frac{x-y}{x+6}\)
b)Tính L = \(\frac{2x+6}{3x-2y}+\frac{2y-6}{4y-x}\)
NH
Những câu hỏi liên quan
cho x-2y=6 tính:
A= x-y/x+6
B= 2x+6/3x-2y + 2y-6/4y-x
Tính:
\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\)
\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(\)Thank you so much!
a.\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\) MTC=6x2y
\(=\frac{3\left(4x-1\right)}{6x^2y}-\frac{2\left(7x-1\right)}{6x^2y}\)
\(=\frac{12x-3-\left(14x-2\right)}{6x^2y}\)
\(=\frac{12x-3-14x+2}{6x^2y}\)
\(=\frac{-2x-1}{6x^2y}=\frac{2\left(-x-1\right)}{6x^2y}=-\frac{x-1}{3x^2y}\)
b.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) MTC= 2x (x + 3)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c.\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)MTC= xy (x+2y).(x-2y)
\(=\frac{2xy\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{xy\left(x+2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{3x^2y-2xy^2+4xy}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{xy\left(3x-2y+4\right)}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
Chọn mk nha!
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Tính giá trị của:
a,L=2x+6/3x-2y + 2y-6/4y-x tại x-2y=6
b,M=(1-z/x).(1-y/z).(1+y/z) (x,y,z ko phải 0) tại x-y-z=0
giải nhanh giúp mình ạ
Ai giúp tui bài toán này với, tui tick cho 4 cái:
Cho x-2y=6. Tính:
A= x-y/x+6 (x+y; x khác (-6)
B= (2x+6/3x-2y)+(2y-6/4y-x) (3x khác 2y; x khác 4y)
Ai giải nhanh nhất, có lời giải tớ tick nhiệt tình nha
ai júp với, tớ T I C K cho mỗi ngày, gấp lắm
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8,Thực hiện phép tínha,frac{5x^2-y^2}{xy}-frac{3x-2y}{y}b,frac{3}{2x+6}-frac{x-6}{2x^2+6x}c,frac{2x}{x^2+2xy}+frac{y}{xy-2y^2}+frac{4}{x^2-4y^2} d,frac{1}{x-y}+frac{3xy}{y^3-x^3}+frac{x-y}{x^2+xy+y^2} e,frac{2x+y}{2x^2-xy}+frac{16x}{y^2-4x^2}+frac{2x-y}{2x^2+xy} f,frac{1}{1-x}+frac{1}{1+x}+frac{2}{1+x^2}+frac{4}{1+x^4}+frac{8}{1+x^8}+frac{16}{1+x^{16}}
Đọc tiếp
8,Thực hiện phép tính
a,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
b,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
d,\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
e,\(\frac{2x+y}{2x^2-xy}+\frac{16x}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
f,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
Thực hiện các phép toán
a/ \(\frac{x+1}{2x+6}\)+ \(\frac{2x+3}{x^2+3x}\)
b/ \(\frac{3}{2x+6}\)- \(\frac{x-6}{2x^2+6x}\)
c/ \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
Cac ban bam' vao` xem them se~ thay' bai` cua minh`
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a)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}\) và 3x-2y+3z
b )\(x=\frac{y}{6}=\frac{z}{3}\) và 2x-3y+4z=-24
c)\(\frac{x}{2}=y=\frac{z}{3}\) và 3x-2y=4z
a, Thiếu đề
b, Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{1}=\frac{y}{6}=\frac{z}{3}=\frac{2x-3y+4z}{2-18+12}=-\frac{24}{-4}=6\)
\(x=6;y=36;z=18\)
c, Ta có : \(3x-2y=4z\Leftrightarrow3x-2y-4z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=\frac{3x-2y-4z}{6-2-12}=\frac{0}{-8}=0\)
\(x=y=z=0\)
b) Đặt \(x=\frac{y}{6}=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=k\\y=6k\\z=3k\end{cases}}\)
Khi đó 2x - 3y + 4z = -24
<=> 2k - 3.6k + 4.3k = -24
=> 2k - 18k + 12k = -24
=> -4k = -24
=> k = 6
=> x = 1 ; y = 36 ; z = 18
c) Đặt \(\frac{x}{2}=y=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=2k\\y=k\\z=3k\end{cases}}\)
Khi đó 3x - 2y = 4z
<=> 3.2k - 2k = 4.3k
=> 6k - 4k = 12k
=> 2k = 12k
=> k = 0
=> x = y = z = 0
b , Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
x/1 = y/6 = z/3 = 2x - 3y + 4z / 2 - 18 + 12 = 24/-4 = 6
=> x = 6
y= 36
z = 18
c, Theo đb , ta có : 3x-2y=4z ,
<=> 3x - 2y - 4z = 0
A/D tính chất dãy tỉ số bằng nhau , ta có :
x/2 = y/1 = z/3 = 3x-2y-4z / 6 -2-12 = 0/-8 = 0
=> x = y =z = 0
Cho \(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\frac{4y-5z}{6}\)và 3x - 2y + 5z = 96 . Tính x , y , z .
1,Thực hiện phép tính
a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
b,\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
c,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
a) \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3x-2x+4x^2-2x-4x^2+4x-4x+4}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{2x\left(x-1\right)}\)
b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)
c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)
\(=\frac{5x^2-y-3x^2+2xy}{xy}\)
\(=\frac{2x^2-y+2xy}{xy}\)
d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}\)
\(=\frac{2x+6}{2x\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{2}{2x}=\frac{1}{x}\)