\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân VTV

\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)

\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)

a: 2x^2y-50xy=2xy(x-25)

b: 5x^2-10x=5x(x-2)

c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)

d: \(x^2-xy+x=x\left(x-y+1\right)\)

e: x(x-y)-2(y-x)

=x(x-y)+2(x-y)

=(x-y)(x+2)

f: 4x^2-4xy-8y^2

=4(x^2-xy-2y^2)

=4(x^2-2xy+xy-2y^2)

=4[x(x-2y)+y(x-2y)]

=4(x-2y)(x+y)

f1: x^2ỹ-y^2+y

=(x-y)(x+y)+(x+y)

=(x+y)(x-y+1)

1. x² + 4y² + x²y² - 8xy + 4 = 0

⇔ x² - 4xy + 4y² + x²y² - 4xy + 4 = 0

⇔ ( x - 2y )² + (xy - 2 )² = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\xy-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\xy=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm2\\y=\pm1\end{matrix}\right.\)

⇒ x + y = 3 hoặc x + y = - 3

mình khuyên bạn nên đưa lên từng câu một thôi chứ bạn đưa lên dài thế này ai nhìn cũng khong muốn làm đâu nha

BẠN HÃY DÙNG Fx ĐỂ GHI CHO DỄ HIỂU NHÉ BẠN

\(\left\{{}\begin{matrix}mx-y=4\\x+my=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx=y+4\\my=-2-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}mxy=y^2+4y\left(y\ne0\right)\\mxy=-2x-x^2\left(x\ne0\right)\end{matrix}\right.\).
Suy ra \(y^2+4y=-2x-x^2\Leftrightarrow x^2+y^2+4y+2x=0\).

Một năm trôi qua ~ . Giờ làm tiếp câu 1 :v

Câu a : \(x\left(x-y\right)+y\left(x-y\right)=x^2-xy+xy-y^2=x^2-y^2\)

Câu b : \(\left(x^2-xy+y^2\right)\left(x+y\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=\left(x^3+y^3\right)-\left(x^3-y^3\right)=x^3+y^3-x^3+y^3=2y^3\)

Câu c : \(7x\left(4y-x\right)+4y\left(y-7x\right)-\left(4y^2-7x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x^2=0\)

Câu d : \(\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\)

\(=4xz+2xy+2yz+y^2+xy-xz-y^2+yz\)

\(3xy+3yz+3xz=3\left(xy+yz+xz\right)\)

Lười làm câu 1 :

Câu 2 :

\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)

\(\Leftrightarrow15x=30\)

\(\Rightarrow x=2\)

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)

a) \(xy+y^2-x-y\)

\(=\left(xy+y^2\right)-\left(x+y\right)\)

\(=y\left(x+y\right)-\left(x+y\right)\)

\(=\left(y-1\right)\left(x+y\right)\)

a) xy +y² - x-y

y(x+y) -(x+y)

(x+y)(y-1)

c) x² - 4x +3

 x² -3x - x - 3

x(x-3) -(x-3)

(x-3)(x-1)

câu 2

ĐỂ phép chia hết thì m+12 = 0 => m = -12

có thể đúng cũng có thể sai ,có j sai hoặc ko đúng ib mk nhé