- \(\frac{x}{3}\)=\(\frac{y}{6}\). Tìm x,y biết:
- x+y=90
- 4x-y=42
- x.y=162
- 2x mũ 2 - y mũ 2= -8
- x-y= 9
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8 tháng 8 2021
6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)
8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2 = -(y-x^2-x) (y+x^2+x)
10, 4x mũ 2 ( x + y ) -x - y = (2x-1) (2x+1) (y+x)
5 tháng 10 2020
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
Y
7 tháng 8 2018
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
30 tháng 6 2016
\(x^2-y^2=1\)
Ta có : \(\left(\frac{x}{5}\right)^2=\left(\frac{y}{4}\right)^2\)
\(=>\frac{x^2}{25}=\frac{y^2}{16}\)
A/d dãy ............
\(\frac{x^2-y^2}{25-16}=\frac{1}{9}=>\frac{x}{5}=\frac{y}{4}=\frac{1}{3}\)
\(=>\frac{x}{5}=\frac{1}{3}=>x=\frac{5}{3}\)
\(=>\frac{y}{4}=\frac{1}{3}=>x=\frac{4}{3}\)
30 tháng 6 2016
\(\frac{x}{5}=\frac{y}{4}\)nên \(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)=> \(\frac{x}{5}=\sqrt{\frac{1}{9}};-\sqrt{\frac{1}{9}}=\frac{1}{3};\frac{-1}{3}\)
=> x = \(\frac{1}{3}.5;\frac{-1}{3}.5=\frac{5}{3};\frac{-5}{3}\)
13 tháng 5 2018
a)<=>
A,=(x+y)(x-y)=x^2-y^2
x=(-1/2)^5:(1/2)^4=-1/2
x^2=1/4
y=8^2/(-2)^5=-2
y^2=4
A=1/4-4=-15/4
2,Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì x+y=90 nên:3k+6k=90
\(\Leftrightarrow\)k(3+6)=90
9k=90
k=90:9=10
Suy ra k=10\(\hept{\begin{cases}x=3.10=30\\y=6.10=60\end{cases}}\)
3,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì 4x-y=42 nên:4.3k-6k=42
\(\Leftrightarrow\) 12k-6k=42
6k=42
k=42:6=7
Suy ra k=7\(\hept{\begin{cases}x=3.7=21\\y=6.7=42\end{cases}}\)
4,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì xy=162 nên:3k.6k=162
\(\Leftrightarrow\)k2.18=162
k2=162:18
k2=9
k=\(\pm\)3
Với k=3\(\hept{\begin{cases}x=3.3=9\\y=6.3=18\end{cases}}\)
Với k=-3\(\hept{\begin{cases}x=3.\left(-3\right)=-9\\y=6.\left(-3\right)=-18\end{cases}}\)
5,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì 2x2-y2=-8 nên:2.(3k)2-(6k)2=-8
\(\Leftrightarrow\)2.9k2-36k2=-8
18k2-36k2=-8
-18k2=-8
k2=-8/-18=4/9
k=\(\pm\)\(\frac{2}{3}\)
Với k=\(\frac{2}{3}\)\(\hept{\begin{cases}x=\frac{2}{3}.3=2\\y=\frac{2}{3}.6=4\end{cases}}\)
Với k=\(\frac{-2}{3}\)\(\hept{\begin{cases}x=\frac{-2}{3}.3=-2\\y=\frac{-2}{3}.6=-4\end{cases}}\)
6,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì x-y=9 nên:3k-6k=9
\(\Leftrightarrow\) -3k=9
k=9:(-3)
k=-3
Suy ra\(\hept{\begin{cases}x=-3.3=-9\\y=-3.6=-18\end{cases}}\)