x=4 3/7 y= 0,5 ạ
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\(x^2-2xy-4z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\) ( 1 )
Thay vào bấm máy tính ta được ( 1 )=19
b) \(3\left(x-3\right)\left(x+7\right)-\left(x-4\right)^2\)
\(=\left(3x-9\right)\left(x+7\right)-\left(x^2-8x+16\right)\)
\(=3x^2+12-63-x^2+8x-16\)
\(=2x^2+20x-79\)
\(=2x^2+20x+50-129\)
\(=2\left(x+5\right)^2-129\)
Thay x vào
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
a) 0,5 x - \(\frac{2}{3}\)x = \(\frac{7}{12}\)
x . ( 0,5 - \(\frac{2}{3}\)) = \(\frac{7}{12}\)
x . \(\frac{-1}{6}\) = \(\frac{7}{12}\)
x = \(\frac{7}{12}\): \(\frac{-1}{6}\)
x = \(\frac{-7}{2}\)
b) x : \(\frac{25}{3}\) = -25
x = -25 . \(\frac{25}{3}\)
x = \(\frac{-625}{3}\)
c) 5,5 x = \(\frac{13}{15}\)
x = \(\frac{13}{15}\): 5,5
x = \(\frac{26}{165}\)
d) (\(\frac{3x}{7}\)+ 1 ) : ( -4) = \(\frac{-1}{28}\)
( \(\frac{3x}{7}\)+ 1 ) = \(\frac{-1}{28}\). ( -4)
\(\frac{3x}{7}\)+ 1 = \(\frac{1}{7}\)
\(\frac{3x}{7}\) = \(\frac{1}{7}\)- 1
\(\frac{3x}{7}\) = \(\frac{-6}{7}\)
=> x = -2
e) y + 30% y = -1,3
y + \(\frac{3}{10}\)y = -1,3
y . ( 1 + \(\frac{3}{10}\)) = -1,3
y . \(\frac{13}{10}\) = -1,3
y = -1,3 : \(\frac{13}{10}\)
y = -1
g) y - 25% y = \(\frac{1}{2}\)
y - \(\frac{1}{4}\)y = \(\frac{1}{2}\)
y . ( 1 - \(\frac{1}{4}\)) = \(\frac{1}{2}\)
y . \(\frac{3}{4}\) = \(\frac{1}{2}\)
y = \(\frac{1}{2}\): \(\frac{3}{4}\)
y = \(\frac{2}{3}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây huhu
a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)
a: Ta có: \(x^2-2xy+y^2-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)
\(=-8000\)
b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)
\(=3x^2+12x-63+x^2-8x+16+48\)
\(=2x^2+4x+1\)
\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)
\(=\dfrac{7}{2}\)
Ta có: \(P = \left( {21{{\rm{x}}^4}{y^5}} \right):\left( {7{{\rm{x}}^3}{y^3}} \right) = \left( {21:7} \right).\left( {{x^4}:{x^3}} \right).\left( {{y^5}:{y^3}} \right) = 3{\rm{x}}{y^2}\)
Thay x = -0,5; y = 2 vào biểu thức \(P = 3{\rm{x}}{y^2}\) ta được:
\(P = 3.\left( { - 0,5} \right){.2^2} = - 6\)
Vậy P = -6 tại x = -0,5; y = 2
a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)
\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)
b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)
c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)
d)\(-3\left(x-2\right)=2x+1\)
\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)
\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)
e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều
\(d,-3\left(x-2\right)=2x+1\)
\(< =>-3x+6=2x+1\)
\(< =>-3x-2x+6-1=0\)
\(< =>5-5x=0\)
\(< =>5\left(1-x\right)=0< =>x=1\)
\(e,\left(x-1\right)^2-4=0\)
\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
a) y : 0,25 + y x 8 - y : 0,5 = 230
y x 4 + y x 8 - y x 2 = 230
y x (4 + 8 - 2) = 230
y x 10 = 230
y = 230 : 10
y = 23
b) \(x\) x 4 + \(x\) x 6 = 30
\(x\) x (4 + 6) = 30
\(x\) x 10 = 30
\(x\) = 30 : 10
\(x\) = 3
a; y :0,25 + y x 8 - y: 0,5 = 230
y x 4 + y x 8 - y x 2 = 230
y x (4 + 8 - 2) = 230
y x 10 = 230
y = 230 : 10
y = 23
b; \(x\) \(\times\) 4 + \(x\) \(\times\) 6 = 30
\(x\) \(\times\) (4 + 6) = 30
\(x\) \(\times\) 10 = 30
\(x\) = 30 : 10
\(x\) = 3
\(x\) = 4\(\dfrac{3}{7}\) = \(\dfrac{31}{7}\)
Thay \(x\) = \(\dfrac{31}{7}\) và y = 0,5 vào biểu thức: \(x^2\) . y2 ta có:
(\(\dfrac{31}{7}\))2 . (0,5)2 = (\(\dfrac{31}{7}\) . 0,5)2 = \(\left(\dfrac{31}{14}\right)^2\) = \(\dfrac{961}{196}\)