Giải pt sau
a)x^2+6=xy-y^2
b) 4x^2+4x+y^2=24
c) 2x+5y=48
d) 10x+y=18
e) 4x^2+5y^2=145
f) ||x-2|+3|=5
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải pt sau
a)x^2+6=xy-y^2
b) 4x^2+4x+y^2=24
c) 2x+5y=48
d) 10x+y=18
e) 4x^2+5y^2=145
f) ||x-2|+3|=5
a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
\(A=\left(x-1\right)^2+2\ge2\)
\(B=-\left(x+2\right)^2+7\le7\)
\(C=2\left(x+1\right)^2+3\ge3\)
\(D=\left(x-1\right)^2+2\left(y+3\right)^2+\left(3z+1\right)^2+4\ge4\)
\(E=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2-\frac{33}{4}\ge-\frac{33}{4}\)
\(F=\left(x-2\right)^2+\left(y+1\right)^2\ge0\)
\(G=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(H=-x^2+7x+74=-\left(x-\frac{7}{2}\right)^2+\frac{345}{4}\le\frac{345}{4}\)
có thể trả lời đầy đủ giúp mình câu b, c, d, h được ko ??????????
a) \(-x-y^2+x^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right).1\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b) \(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-5\right)\)
c) \(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
d) \(5x^3-5x^2y-10x^2+10xy\)
\(=5x\left(x^2-xy-2x+2y\right)\)
\(=5x\left[x\left(x-y\right)-2\left(x-y\right)\right]\)
\(=5x\left(x-y\right)\left(x-2\right)\)
e) \(27x^3-8y^3\)
\(=\left(3x\right)^3-\left(2y\right)^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x2y+\left(2y\right)^2\right]\)
\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
f) \(x^2-y^2-x-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
g) \(x^2-y^2-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-y^2\)
\(=\left(x-y\right)^2-y^2\)
\(=\left(x-y-y\right)\left(x-y+y\right)\)
\(=\left(x-y^2\right)x\)
h) \(x^2-y^2+4-4x\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x^2-2.2x+2^2\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
i) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)