\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)

\(\Leftrightarrow x=1\)

( 198 : 18 - 143 : 13 ) x ( 16996 - 1110 : 30 x 305 )

= ( 11 - 11 ) x ( 16996 - 1110 : 30 x 305 )

= 0 x ( 16996 - 1110 : 30 x 305 )

= 0

Ta có :\(\frac{1}{a}+\frac{1}{b}=\frac{4}{ab}-1\)

\(\Rightarrow\frac{a+b}{ab}=\frac{4}{ab}-1\)

=> \(\frac{a+b-4}{ab}=-1\)

=> a + b - 4 = -ab

=> a + b - 4 + ab = 0

=> a(b + 1) + b + 1 - 5 = 0

=> (a + 1)(b + 1) = 5

Vì \(a;b\inℤ\Rightarrow\hept{\begin{cases}a+1\inℤ\\b+1\inℤ\end{cases}}\)

Khi đó 5 = 1.5 = (-1).(-5)

Lập bảng xét các trường hợp

Vậy các cặp (a;b) nguyên thỏa mãn là (-6 ; -2) ; (-2 ; -6)

\(\frac{1}{a}+\frac{1}{b}=\frac{4}{ab}-1\)( ĐKXĐ : \(a,b\ne0\)) ( Bạn Xyz nhớ bổ sung thêm ĐKXĐ ạ )

\(\Leftrightarrow\frac{b}{ab}+\frac{a}{ab}=\frac{4}{ab}-\frac{ab}{ab}\)

\(\Leftrightarrow\frac{b}{ab}+\frac{a}{ab}-\frac{4}{ab}+\frac{ab}{ab}=0\)

\(\Leftrightarrow\frac{b+a-4+ab}{ab}=0\)

\(\Leftrightarrow b+a-4+ab=0\)

\(\Leftrightarrow b+a-5+1+ab=0\)

\(\Leftrightarrow a\left(b+1\right)+1\left(b+1\right)=5\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)=5\)

Ta có bảng sau :

Theo ĐKXĐ => Các cặp  ( x; y ) thỏa mãn là : ( -2 ; -6 ) ; ( -6 ; -2 )

Áp dụng HĐT a² - b² = ( a + b )( a - b ) ta có :

50² - 49² + 48² - 47² + ... + 2² - 1²

= ( 50² - 49² ) + ( 48² - 47² ) + ... + ( 2² - 1² )

= ( 50 + 49 )( 50 - 49 ) + ( 48 + 47 )( 48 - 47 ) + ... + ( 2 + 1 )( 2 - 1 )

= 99.1 + 95.1 + ... 3.1

= 99 + 95 + ... + 3

= \(\frac{\left(99+3\right)\left[\left(99-3\right):4+1\right]}{2}\)

= 1275

Áp dụng tính chất a² - b² = a² + ab - ab - b² = a(a + b) - b(a + b) = (a - b)(a + b)

Khi đó (50² - 49²) + (48² - 47²) + ... + (2² - 1²)

= (50 + 49)(50 - 49) + (48 + 47)(48 - 47) + .... + (2 + 1)(2 - 1)

= 50 + 49 + 48 + 47 + ... + 2 + 1

= 50(50 + 1) : 2 = 1275 

\(x\in\left(2;+\infty\right)\)

Ta có: \(x^2+y^2-4x=6z-2y-z^2-14\)

\(x^2+y^2-4x-6z+2y+z^2+14=0\)

\(\left(x^2-4x+2^2\right)+\left(y^2+2y+1\right)+\left(z^2-6z+3^2\right)=0\)

\(\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)

\(\cdot\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

\(\cdot\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\)

\(\left(z-3\right)^2=0\Rightarrow z-3=0\Rightarrow z=3\)

hok tốt!

Ta có x² + y² - 4x = 6z - 2y - z² - 14

=> x² + y² - 4x - 6z + 2y + z² + 14 = 0

=> (x² - 4x + 4) + (y² + 2y + 1) + (z² - 6z + 9) = 0

=> (x - 2)² + (y + 1)² + (z - 3)² = 0

Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y+1=0\\z-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)

Vậy x = 2 ; y = - 1 ; z = 3

???????

CHẢ HIỂU 

Ta có: x + y = a + b

\(\Rightarrow\left(x+y\right)^2=\left(a+b\right)^2\)

\(\Rightarrow x^2+y^2=a^2+b^2\)(đpcm)

đề hơi sai!!:))

hok tốt!

Ta có : x + y = a + b (1)

=> (x + y)³ = (a + b)³

=> x³ + y³ + 3x²y + 3y²x = a³ + b³ + 3ab² + 3a²b

=> 3x²y + 3y²x = 3ab² + 3a²b

=> 3xy(x + y) = 3ab(a + b)

=> 3xy = 3ab

=> xy = ab

Từ (1) => (x + y)² = (a + b)²

=> x² + y² + 2xy = a² + b² + 2ab

=> x² + y² = a² + b² (Vì xy = ab => 2xy = 2ab) (đpcm)