Các câu hỏi tương tự
Tìm các cặp số nguyên (x;y) thõa mãn
\(\text{|}x-5\text{|}+\text{|}1-x\text{|}=\frac{12}{\text{|}y+1\text{|}+3}\)
Cho x + 3y - 2z = 36 . Tìm x,y,z biết :
a)\(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b)\(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{3};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
Cho x + 3y - 2z = 36. Tìm x,y,z biết
a) \(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b) \(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{\text{3}};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
\(\frac{y+\text{z}+1}{x}=\frac{x+\text{z}+2}{y}\frac{x+y-3}{\text{z}}=\frac{1}{x+y+\text{z}}\)
\(Tìm\text{ }x,y\text{ }\text{ }biết:\)
\(x-\frac{y}{3}=x+\frac{y}{13}=\frac{x.y}{200}\)
Tìm x biếta,left(frac{4}{5}right)^{2x+7}text{}frac{625}{256}b,frac{7^{x+2}+7^{x+1}+7^x}{57}text{}frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}c,left(4x-3right)^4text{}left(4x-3right)^2d,frac{2x+3}{5x+2}text{}frac{4x+5}{10x+2}e,frac{3x-1}{40-5x}text{}frac{2x-3x}{5x-34}f,frac{15}{x-9}text{}frac{20}{y-12}text{}frac{40}{z-2x} và xytext{}1200
Đọc tiếp
Tìm x biết
a,\(\left(\frac{4}{5}\right)^{2x+7}\text{=}\frac{625}{256}\)
b,\(\frac{7^{x+2}+7^{x+1}+7^x}{57}\text{=}\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
c,\(\left(4x-3\right)^4\text{=}\left(4x-3\right)^2\)
d,\(\frac{2x+3}{5x+2}\text{=}\frac{4x+5}{10x+2}\)
e,\(\frac{3x-1}{40-5x}\text{=}\frac{2x-3x}{5x-34}\)
f,\(\frac{15}{x-9}\text{=}\frac{20}{y-12}\text{=}\frac{40}{z-2x}\) và \(xy\text{=}1200\)
Cho \(\dfrac{\text{x}}{\text{2}}=\dfrac{\text{y}}{\text{3}}=\dfrac{\text{z}}{\text{5}}\). Tìm x,y,z biết
a) x + y + z = 40
b) x - 3y + 2z = 9
c) x -y + z = 28
d) 3x + 2y = 24
1 . Tìm x , y biết : \(\frac{x+y}{16}=\frac{xy}{17}=\frac{x-y}{18}\)
2 . Một số chính phương có dạng abcd . Biết ab - cd =1 . . Hãy tìm số abcd
cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)