450,3 - ( 21,5 + y : 1/2 ) x 15 = 0
A.x * 15 - 12,3 * x =81
B.x : 2,5 - 4,4 =3,8
C.37 - x + 21,5 = 35,8
CẢM ƠN TRƯỚC
\(a,x\times15-12,3\times x=81\)
\(x\times\left(15-12,3\right)=81\)
\(x\times2,7=81\)
\(x=80:2,7\)
\(x=30.\)
\(b,x:2,5-4,4=3,8\)
\(x:2,5=3,8+4,4\)
\(x:2,5=8,2\)
\(x=8,2\times2,5\)
\(x=20,5.\)
\(c,37-x+21,5=35,8\)
\(37-x=35,8-21,5\)
\(37-x=14,3\)
\(x=37-14,3\)
\(x=22,7.\)
Cho x,y>0,x+y=1.CM:`A=(x+1/x)^2+(y+1/y)^2>=25/2`
`A=x^2+1/x^2+2+y^2+1/y^2+2`
`=x^2+y^2+1/x^2+1/y^2+4`
`=(x^2+1/(16x^2))+(y^2+1/(16y^2))+4+15/16(1/x^2+1/y^2)`
Áp dụng BĐt cosi và `1/a^2+1/b^2>=8/(a+b)^2`
`=>A>=1/2+1/2+4+15/16(8/(x+y)^2)`
`<=>A>=5+15/2=25/2`
Dấu "=" `<=>x=y=1/2`
Không làm theo cách sau:
Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
2. Tính P=(1+x/y)*(1+z/x)*(1+z/y). Biết x+y+z=0 và x,y,z #0
3. Tính Q= 5.y^10-y^15+2016. Biết (x+1)^2016+(y-1)^2018=0
2. Tính P=(1+x/y)*(1+z/x)*(1+z/y). Biết x+y+z=0 và x,y,z #0
3. Tính Q= 5.y^10-y^15+2016. Biết (x+1)^2016+(y-1)^2018=0
A)1/4+x+x+21,5=36,5
B)3/5×x+x×2+40=53
\(a.\frac{1}{4}+x+x+21,5=36,5\)
\(0,25+21,5+2x=36,5\)\(\)
\(21,75+2x\) \(=36,5\)
\(2x\) \(=36,5-21,75\)
\(x\) \(=14,72\div2=\frac{59}{8}\)
\(b.\frac{3}{5}\times x+x\times2+40=53\)
\(x\times(\frac{3}{5}+2)\) \(=53-40\)
\(x\times\frac{13}{5}\) \(=13\)
\(x\) \(=13\div\frac{13}{5}\)
\(x\) \(=5\)
Học tốt nhé
Giải các phương trình sau :
v) x+1 / 2009 + x+3 / 2007 = x+5 / 2005 + x+7 / 1993
x) 392- x / 32 + 390 - x / 34 + 388 - x / 36 + 386 - x / 38 + 384 - x / 40 = -
y ) x - 15 / 23 + x - 23 / 15 - 2 =0
a ) y(y^2 - 1) -y^2 - 5y+6 = 0
b ) y( y-1/2 )(2y+5) = 0
m ) y^2 - y -12 = 0
n ) x^2 + 2x + 7 = 0
o ) y^3 - y^2 - 21y +45 = 0
p ) 2y^3 - 5y^2 + 8y - 3 = 0
q ) ( y+3 )^2 + (y+5)^2 = 0
\(\frac{ }{ }\)\(\frac{ }{ }\)
Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !
Tìm x,y,z:
a)|1-x|+|y-2/3|+|x+z|=0
b)|1/4-x|+|x+y+z|+|2/3+y|=0
c)|15/32-x|+4/25-y|+|z-14/31|=0
a) \(\left|1-x\right|+\left|y-\frac{2}{3}\right|+\left|x+z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}1-x=0\\y-\frac{2}{3}=0\\x+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-0=1\\y=0+\frac{2}{3}=\frac{2}{3}\\z=0-1=-1\end{cases}}}\)
Vậy \(x=1,y=\frac{2}{3},z=-1\)
b) \(\left|\frac{1}{4}-x\right|+\left|x+y+z\right|+\left|\frac{2}{3}+y\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}-x=0\\x+y+z=0\\\frac{2}{3}+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-0=\frac{1}{4}\\x+y+z=0\\y=0+\frac{2}{3}=\frac{2}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\z=0-\frac{1}{4}-\frac{2}{3}=\frac{-11}{12}\\y=\frac{2}{3}\end{cases}}}\)
Vậy \(x=\frac{1}{4},y=\frac{-11}{12},z=\frac{2}{3}\)
giúp e với ạ..!
y x 54=351
(21,5%+41,5%):7
ạ
a . y= 351 : 54 = 6,5
b. (0,215+0,415):7 = 0,63 : 7 = 0,09
1. tìm x |x| + x = 0 x + |x| = 2x x/|x| = -1 |3x-2| = x |x-2| = 2x + 1
2.tìm x , y, z thuộc Q |x +19/5| + | y + 1890/1975| + | z- 2004 | = 0 |x- 1/2| + | y+ 3/2 | + | x -y -z -1/2 | = 0
|15/32 - x | + |4/25 - y| + | z- 13/31| < 0
Câu 2:
a: Ta có: \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{378}{395}\\z=2004\end{matrix}\right.\)
b: \(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{3}{2}\right|+\left|x-y-z-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{3}{2}=0\\x-y-z-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{3}{2}\\z=\dfrac{3}{2}\end{matrix}\right.\)