Yêu cầu:tìm x

Với x \(\inℕ\)ta có : 1 + x(x + 2) = x² + 2x + 1 = (x + 1)²

Ta có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\)

=> \(\left(\frac{1+1.3}{1.3}\right)\left(\frac{2.4+1}{2.4}\right)\left(\frac{3.5+1}{3.5}\right)....\left(\frac{x\left(x+2\right)+1}{x\left(x+2\right)}\right)=\frac{31}{16}\)

=> \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{\left(x+1\right)^2}{x\left(x+2\right)}=\frac{31}{16}\)

=> \(\frac{2^2.3^2.4^2...\left(x+1\right)^2}{1.3.2.4.3.5...x\left(x+2\right)}=\frac{31}{16}\)

=> \(\frac{\left(2.3.4...\left(x+1\right)\right).\left(2.3.4....\left(x+1\right)\right)}{\left(1.2.3...x\right)\left(3.4.5...\left(x+2\right)\right)}=\frac{31}{16}\)

=> \(\frac{\left(x+1\right).2}{x+2}=\frac{31}{16}\)

=> 31(x + 2) = (x + 1).2.16

=> 31x + 62 = (x + 1).32

=> 31x + 62 = 32x + 32

=> 32x - 31x = 62 - 32

=> x = 30

Vậy x = 30

Giải phương trình nhé!

Ta có x³ - y³ + z³ + 3xyz

= (x - y)(x² + xy + y²) + z³ + 3xyz

= (x - y)³ + 3xy(x - y) + z³ + 3xyz

= [(x - y)³ + z³] + (3xy(x - y) + 3xyz)

= (x - y + z)[(x - y)² - (x - y).z + z²] + 3xy(x - y + z)

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

= (x - y + z)(x² + y² + z² + xy + yz - zx)

Lại có (x + y)² + (y + z)² + (z - x)²

= x² + y² + 2xy + y² + z² + 2yz + z² - 2xz + x²

= 2x² + 2y² + 2z² + 2xy + 2yz - 2zx

= 2(x² + y² + z² + xy + yz - zx)

Khi đó \(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}=\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-zx\right)}{2\left(x^2+y^2+z^2+xy+yz-zx\right)}\)

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

Đặt \(A=\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

Xét tử, ta có: 

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

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

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

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

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

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

Xét mẫu, ta có: 

\(\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\)

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

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

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

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

sai đề rồi nhé , đề phải là :

\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\frac{\left(x-y\right)^3+3xy.\left(x-y\right)+z^3+3xyz}{x^2+2xy+y^2+y^2+2yz+z^2+z^2-2xz+x^2}\)

\(=\frac{\left(x-y+z\right).\left[\left(x-y\right)^2-\left(x-y\right).z+z^2\right]+3xy.\left(x-y+z\right)}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)

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

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

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

A B C O Q P F E D

Từ A kẻ đường thẳng // BC cắt BO, CO kéo dài tại P và Q

Theo định lý Thales ta có: \(\frac{DB}{DC}=\frac{AP}{AQ},\frac{EC}{EA}=\frac{BC}{AP},\frac{FA}{FB}=\frac{AQ}{BC}\)

Nhân 3 đẳng thức vs nhau ta đc: 

\(\frac{DB}{DC}.\frac{EC}{EA}.\frac{FA}{FB}=\frac{AP}{AQ}.\frac{BC}{AP}.\frac{AQ}{BC}=1\) ( ĐPCM)

Um vậy để tôi nghĩ tiếp 

a)Ta có : \(P=\frac{x^2}{x-1}< 1\)

\(\Leftrightarrow\frac{x^2}{x-1}-1< 0\)

\(\Leftrightarrow\frac{x^2-x+1}{x-1}< 0\)

Ta lại có : \(x^2-x+1=\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}\)\(=\left(x-\frac{1}{2}\right)+\frac{3}{4}>0\forall x\)

\(\Rightarrow P< 1\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Vậy \(\hept{\begin{cases}x< 1\\x\ne0\end{cases}}\)thì \(P< 1\)

b) Đề có sai không ạ ? Nếu \(x\ge1\)thì có thể ra kết quả 

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)

Ta có tử bằng:2x³-7x²-12x+45

                    =(2x³-6x²)-(x²-3x)-(15x-45)

                    =2x²(x-3)-x(x-3)-15(x-3)

                    =(x-3)(2x²-x-15)

                    =(x-3)(2x²-6x+5x-15)

                   =(x-3)²(2x+5)                   (1)

Ta có mẫu bằng:3x³-19x²+33x-9

                        =(3x³-x²)-(19x²-6x)+(27x-9)

                        =x²(3x-1)-6x(3x-1)+9(3x-1)

                        =(3x-1)(x²-6x+9)

                        =(3x-1)(x-3)²                (2)

Thay (1) và (2) vào phân thức ,ta có:

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}=\frac{2x+5}{3x-1}\)