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1) \(3x=2y\)và \(\left(x+y\right)^3-\left(x-y\right)^3=126\)
Có: \(3x=2y\)=> \(\frac{x}{2}=\frac{y}{3}=\frac{x-y}{2-3}=\frac{x+y}{2+3}\)
=> \(\frac{x+y}{5}=\frac{x-y}{-1}\)
=> \(\frac{\left(x+y\right)^3}{5^3}=\frac{\left(x-y\right)^3}{\left(-1\right)^3}=\frac{\left(x+y\right)^3-\left(x-y\right)^3}{5^3-\left(-1\right)^3}=\frac{126}{126}=1\)
=> \(\hept{\begin{cases}\frac{\left(x+y\right)^3}{5^3}=1\\\frac{\left(x-y\right)^3}{\left(-1\right)^3}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=5\\x-y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5+\left(-1\right)}{2}=2\\y=\frac{5-\left(-1\right)}{2}=3\end{cases}}\)
Vậy:...
2) Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{2}=\frac{z}{-3}=\frac{2x-3y+4z}{2.3-3.2+4.\left(-3\right)}=\frac{48}{-12}=-4\)
=>
\(\frac{x}{3}=-4\Rightarrow x=-12\)
\(\frac{y}{2}=-4\Rightarrow y=-8\)
\(\frac{z}{-3}=-4\Rightarrow z=12\)
Vậy:...
`(x+1/4)^2=x^2+1/2 x + 1/16`
`(3x^2-2y)^3=27x^6-54x^4y+36x^2y^2-8y^3`
`(2/3 x^2 -1/2 y)^3=8/27 x^6 - 2/3 x^4y+1/2 x^2y^2 - 1/8 y^3`
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
Bài 1 :
\(A=x^2-2xy^2+y^4=\left(x-y^2\right)^2=-\left(y^2-x\right)^2\)
Mà \(B=-\left(y^2-x\right)^2\)
Nên ta có : đpcm
Bài 2
Đặt \(\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
TH1 : x = -1
TH2 : x = 2
TH3 : x = 1/2
Bài 4 :
a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)
b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)
c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)
d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)
3x = 2y
x = 2y/3
(x+y)^3 - (x-y)^3 = 126
(2y/3 + y)^3 - (2y/3 - y)^3 = 126
(5y/3)^3 - (-y/3)^3 = 126
125y3/27 - (-y3/27) = 126
125y3/27 + y3/27 = 126
126y3/27 = 126
126y3 = 126 . 27
=> y3 = 27
=> y = 3
3x = 2y
3x = 6
=> x = 2
đầy đủ lắm rùi đó chế mik nha chế