*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

\(a,4\left(2-x\right)^2+xy-2y\)

\(=4\left(2-x\right)^2-y\left(2-x\right)\)

\(=4-y\left(2-x\right)^2\left(2-x\right)\)

\(=\left(2-x\right)\left[\left(2-x\right)4-y\right]\)

\(=\left(2-x\right)\left(4x-8+y\right)\)

\(c,x^3+y^3+z^3-3xyz\)

\(=x^3+y^3+z^3+3x^2y-3x^2y+3xy^2-3xy^2-3xyz\)

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

\(=\left(x+y\right)^3-3xy\left(x+1\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz\)

\(=\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

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

a) 4(2 - x)² + xy - 2y = 4(x - 2)² + y(x - 2) = (4x - 8 + y)(x - 2)

b) 2(x - 1)³ - 5(x - 1)² - (x - 1) = (x - 1)[2(x - 1)² - 5(x - 1) - 1]

= (x - 1)(2x² - 4x + 2 - 5x + 5 - 1) = (x - 1)(2x² - 9x + 6)

c) x³ + y³ + z³ - 3xyz = (x + y)(x² - xy + y²) + z³ - 3xyz

= (x + y)³ + z³ - 3xy(x + y) - 3xyz = (x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z)

= (x + y + z)(x² + y² + z² - xz - yz + 2xy - 3xy) = (x + y + z)(x² + y² + z² - xy - yz - xz)

Thanks bạn nhen!

Câu 1: \(x^2+5x=2\left(x+5\right)\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

=>x=-5 hoặc x=2

Câu 2: 

Ta có: \(5x^4y^6⋮4x^2y^n\)

=>6-n>=0

hay n<=6

\(xy+x+y=4\)

\(\Leftrightarrow xy+x+y+1=4+1\)

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=5\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=5\)

\(\Leftrightarrow x+1;y+1\inƯ\left(5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=1\\y+1=5\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=5\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-1\\y+1=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-5\\y+1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

Dự đoán dấu bằng xảy ra khi \(x=y=z=2\), áp dụng BĐT AM-GM ta có:

\(8^x+8^x+64\ge3\sqrt[3]{8^x\cdot8^x\cdot64}=12\cdot4^x\)

\(8^y+8^y+64\ge3\sqrt[3]{8^y\cdot8^y\cdot64}=12\cdot4^y\)

\(8^z+8^z+64\ge3\sqrt[3]{8^z\cdot8^z\cdot64}=12\cdot4^z\)

Suy ra \(2\left(8^x+8^y+8^z\right)+3\cdot64\ge12\left(4^x+4^y+4^z\right)\left(1\right)\)

Theo giả thiết ta có:

\(8^x+8^y+8^z\ge3\sqrt[3]{8^{x+y+z}}=3\sqrt[3]{8^6}=3\cdot64\left(2\right)\)

Cộng (1) với (2) theo vế ta có:

\(3\left(8^x+8^y+8^z\right)\ge12\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)

x - 1/3 = 1/4 . x 

-1/3 = 1/4 .x -x 

-1/3 = x . [1/4 -1 ]

-1/3 = x . -3/4

x . -3/4 = -1/3 

x = -1/3 ; -3/4

x = 4/9

ta có \(\left(n^2-n+1\right)+\left(n^2+n+1\right)\\ =n^2-n+1+n^2+n+1\\ =2n^2+2\)

=>\(n\in\left\{n\in N\right\}112\le n\ge123\)

bài này mk k bt cách trình bày nhưng kết quả hình như là 15 đó bạn....

#)Giải :

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x+y+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(x+y+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}\)

\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\)

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(1\right)\\x+y+2=2y\left(2\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y-3=2z\left(3\right)\\x+y+z=\frac{1}{2}\left(4\right)\end{cases}}\)

Ta có : 

\(\left(\cdot\right)x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\) Thay \(\left(1\right)\) vào ta được :

\(\frac{1}{2}-x+1=2x\Rightarrow x=\frac{1}{2}\)

\(\left(\cdot\right)x+y+z=\frac{1}{2}\Rightarrow x+z=\frac{1}{2}-y\) Thay \(\left(2\right)\) vào ta được :

\(\frac{1}{2}-y+2=2y\Rightarrow y=\frac{5}{6}\)

\(\left(\cdot\right)x+y+z=\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Rightarrow z=\frac{-5}{6}\)

Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}\)

phải có 2 trường hợp

TH1 x+y+x=0

TH2 x+y+z khác 0 chứ