Bài 1: Tìm x,y biết :
a) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
b)\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
Bài 2: Timd x,y,z biết:
a)\(5x=-10y=6zvàxyz=-30000\)
b)\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}\)
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a) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\)\(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(\Leftrightarrow20x^2-20x^2+4x+30x-25x-8x=10-6\)
\(\Leftrightarrow x=4\)
b) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow15x^2-15x^2-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=7\)
a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(4x+30x-25x-8x=10-6\)
\(x=4\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2.\left(2x+3\right)-\left(4x+5\right)}{2.\left(5x+2\right)-\left(10x+2\right)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)
Suy ra:
\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow2.\left(2x+3\right)=1.\left(5x+2\right)\Rightarrow4x+6=5x+2\)
\(\Rightarrow x=4\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
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)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(\Leftrightarrow4x+30x-25x-8x=10-6\)
\(\Leftrightarrow1x=4\)
\(\Leftrightarrow x=4:1\)
\(\Leftrightarrow x=4\)
b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)
b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)