tìm các số x,y biết \(\dfrac{2+6y}{24}=\dfrac{2+10y}{10x}=\dfrac{2+14y}{8x}\)
tìm số tự nhiên cs 3 chữ số \(\overline{xyz}\) biết : \(\dfrac{x^2}{4}\) =\(\dfrac{y^2}{9}\)=\(\dfrac{z^2}{25}\) và x-y+z =4
Ta có: \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}=\dfrac{x^2}{2^2}=\dfrac{y^2}{3^2}=\dfrac{z^2}{5^2}\rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
`x/2=y/3=z/5=(x-y+z)/(2-3+5)=4/4=1`
`-> x/2=y/3=z/5=1`
`-> x=2*1=2, y=3*1=3, z=5*1=5`
=>x/2=y/3=z/5 và x-y+z=4
Áp dụng tính chất của DTSBN, ta được:
x/2=y/3=z/5=(x-y+z)/(2-3+5)=4/4=1
=>x=2; y=3; z=5
Ta có: \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)
\(\Rightarrow\dfrac{x}{2}=1\Rightarrow x=2\)
\(\dfrac{y}{3}=1\Rightarrow y=3\)
\(\dfrac{z}{5}=1\Rightarrow z=5\)
Vậy x =2; y =3; z =5
Tìm x,y,z biết\(\dfrac{7y-8x}{10}=\dfrac{10x-7z}{8}=\dfrac{8z-10y}{7}\)
và 2x-3y+5z=1200
mk mới làm 1 ít nhưng đến đó lại bí
ta có \(\dfrac{7y-8x}{10}=\dfrac{10x-7z}{8}=\dfrac{8z-10y}{7}\)
=\(z.\left(\dfrac{7y-8x}{z.10}\right)=y.\left(\dfrac{10x-7z}{y.8}\right)=x.\left(\dfrac{8z-10y}{x.7}\right)\)
=\(\dfrac{7.z.y}{z.10}\)???????
Giúp mk nha!
bài này bn dùng dảy tỉ số bằng nhau nha ! bn xem lại đề hộ mk nhé
Tính :
a)\(\dfrac{6x-3}{5x^2+x}.\dfrac{25x^2+10x+1}{1-8x^3}\)
b)\(\dfrac{3x^2-x}{x^2-1}.\dfrac{1-x^4}{\left(1-3x\right)^3}\)
c)\(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}\)
d) \(\dfrac{5x^2-10xy+5y^2}{2x^2-2xy+2y^2}:\dfrac{8x-8y}{x^3+10y^3}\)
Tìm x , y ,z :
a, \(\dfrac{x+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y và \(2x^2-y^2=-28\)
c, \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
d, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
a)\(\dfrac{3x}{5x+5y}-\dfrac{x}{10x-10y}\)
b)\(\dfrac{7}{8x^2-18}+\dfrac{1}{2x^2+3x}-\dfrac{1}{4x-6}\)
c)\(\dfrac{5}{x+1}-\dfrac{10}{x-x^2+1}-\dfrac{15}{x^3+1}\)
a: \(=\dfrac{3x}{5\left(x+y\right)}-\dfrac{x}{10\left(x-y\right)}\)
\(=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x-y\right)\cdot\left(x+y\right)}\)
\(=\dfrac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}=\dfrac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)
b: \(=\dfrac{7}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{1}{x\left(2x+3\right)}-\dfrac{1}{2\left(2x-3\right)}\)
\(=\dfrac{7x+2\left(2x-3\right)-x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{7x+4x-6-2x^2-3x}{2x\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{-2x^2-6}{2x\left(2x+3\right)\left(2x-3\right)}=\dfrac{-x^2-3}{x\left(2x+3\right)\left(2x-3\right)}\)
c: \(=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)
Bài 1: tìm đạo hàm của các hàm số sau
1. y=6x2 -\(\dfrac{4}{x}\)+1
2. y=\(\dfrac{2x+1}{-x+1}\)
3. y= \(\sqrt{x^2-3x+4}\)
4. y=\(\dfrac{\left(x^2-1\right)\left(x+3\right)}{x-4}\)
5. y=\(\dfrac{1}{2x^2-3x+5}\)
6. y=(x+1)\(\sqrt{x^2-1}\)
1.
\(y'=12x+\dfrac{4}{x^2}\)
2.
\(y'=\dfrac{3}{\left(-x+1\right)^2}\)
3.
\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)
4.
\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)
\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)
5.
\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)
6.
\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)
1. Tìm mã và min
P=x+y-17
biết: \(x^2+2xy-14y-10x+3y^2+27=0\)
2. Cho ab>4
Tim min: \(M=\dfrac{a^2}{b-4}+\dfrac{b^2}{a-4}\)
1.
Đặt \(x+y=a\Rightarrow y=a-x\)
\(\Rightarrow x^2+2x\left(a-x\right)-14\left(a-x\right)-10x+3\left(a-x\right)^2+27=0\)
\(\Leftrightarrow2x^2-4\left(a+1\right)x+3a^2-10a+27=0\)
\(\Delta'=4\left(a+1\right)^2-2\left(3a^2-10a+27\right)\ge0\)
\(\Leftrightarrow-a^2+14a-25\ge0\)
\(\Rightarrow7-2\sqrt{6}\le a\le7+2\sqrt{6}\)
\(\Rightarrow-10-2\sqrt{6}\le P\le-10+2\sqrt{6}\)
2. Chắc đề là \(a;b>0\) (đảm bảo mẫu dương) chứ ko phải \(a.b>4\)
\(M\ge\dfrac{\left(a+b\right)^2}{a+b-8}=\dfrac{\left(a+b-8+8\right)^2}{a+b-8}=\dfrac{\left(a+b-8\right)^2+16\left(a+b-8\right)+64}{a+b-8}\)
\(M\ge a+b-8+\dfrac{64}{a+b-8}+16\ge2\sqrt{\dfrac{64\left(a+b-8\right)}{a+b-8}}+16=32\)
Dấu "=" xảy ra khi \(a=b=8\)
Giải các phương trình sau :
1, 2x (x + 2)\(^2\) - 8x\(^2\) = 2 (x - 2) (x\(^2\) + 2x +4)
2, 8x (x +3 )\(^2\) - 3x = ( x +2)\(^3\) +1
3, \(\dfrac{10x+3}{12}\) = 1 + \(\dfrac{6+8x}{9}\)
4,\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
5,\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
Giúp mk nha mấy bạn mk đg cần gấp nên mong mấy bạn giúp đỡ
timf GTLN của \(\dfrac{y^2-14y-1}{y^2-4y+4}-y^2-6y\)
\(\dfrac{y^2-14y-1}{y^2-4y+4}-y^2-6y\)
\(\Leftrightarrow\dfrac{y^2-14y-1}{y^2-4y+4}-\dfrac{\left(y^2+6y\right)\left(y^2-4y+4\right)}{y^2-4y+4}\)
\(\Rightarrow y^2-14y-1-\left(y^2+6y\right)\left(y^2-4y+4\right)\)
\(\Rightarrow\)y2-14y-1-(y4-4y3+4y2+6y3-24y2+24y)
\(\Rightarrow\)y2-14y-1-y4+4y3-4y2-6y3+24y2-24y
\(\Rightarrow\)-y4-2y3+21y2-38y-1