Ta có: \(x^3+y^3+z^2=3xyz+1\)

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

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

   \(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(zx+zy\right)-3xy\right]=1\)

   \(\Leftrightarrow\left(x+y+z\right)\left[x^2+y^2+z^2+2xy+2yz+2zx-3xy-3yz-3zx\right]=1\)

   \(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=1\)

Đến đây các bạn tự giải nhé ^_^

Bn gì ơi, đây kh pk mk nhờ bn giải hộ, mk nổi hứng đăng câu hỏi lên thôi nên lm hết đi nhá

a) \(\frac{1}{2}+\left(5x-9\right)>\frac{6-5x}{7}+12\)

<=> \(\frac{7}{14}+\frac{14\left(5x-9\right)}{14}>\frac{2\left(6-5x\right)}{14}+\frac{168}{14}\)

<=> \(\frac{7}{14}+\frac{70x-126}{14}>\frac{12-10x}{14}+\frac{168}{14}\)

<=> 7 + 70x - 126 > 12 - 10x + 168

<=> 70x + 10x > 12 + 168 - 7 + 126

<=> 80x > 299

<=> x > 299/80 

b) \(\frac{3x-5}{6}-4x+\frac{2}{5}>\frac{2+5x}{3}\)

\(\Leftrightarrow\frac{5\left(3x-5\right)}{30}-\frac{120x}{30}+\frac{12}{30}>\frac{10\left(2+5x\right)}{30}\)

\(\Leftrightarrow\frac{15x-25}{30}-\frac{120x}{30}+\frac{12}{30}>\frac{20+50x}{30}\)

<=> 15x - 25 - 120x + 12 > 20 + 50x

<=> 15x - 120x - 50x > 20 + 25 - 12

<=> -155x > 33

<=> x < -33/155

Xin phép bỏ biểu diễn trên trục :))

a) \(2x-1< 2\left(x-1\right)\)

\(\Leftrightarrow2x-1< 2x-2\)

\(\Leftrightarrow2x-2x< 1-2\)

\(0x< -1\)( vô lí )

Vậy bất phương trình vô nghiệm.

b) \(\frac{x-1}{3}-\frac{2+3x}{4}>\frac{1}{6}\)

\(\Leftrightarrow\frac{4\left(x-1\right)-3\left(2+3x\right)}{12}>\frac{2}{12}\)

\(\Leftrightarrow4x-4-6-9x>2\)

\(\Leftrightarrow-5x-10>2\)

\(\Leftrightarrow-5x>12\)

\(\Leftrightarrow x< \frac{-12}{5}\)

Vậy...........

            Bài làm :

Ta có :

\(183^2-197.203+17^2+2.183.17\)

\(=\left(183^2+2.183.17+17^2\right)-\left(197.203\right)\)

\(=\left(183+17\right)^2-\left(200-3\right).\left(200+3\right)\)

\(=200^2-\left(200^2-9\right)\)

\(=200^2-200^2+9\)

\(=9\)

183² - 197.203 + 17² + 2.183.17

= (183² + 2.183.17 + 17²) - 197.203

= (183² + 2.183.17 + 17²) - (200 - 3).(200 + 3)

= (183 + 17)² - (200² - 9²)

= 200² - 200² + 9²

= 9²

= 81

Sử dụng BĐT Cauchy Schwarz ta dễ có:

\(P=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)

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

\(\ge\frac{\left(x+y\right)^2}{x+y-2}\)

Ta cần chứng minh: \(\frac{\left(x+y\right)^2}{x+y-2}\ge8\)

\(\Leftrightarrow\left(x+y\right)^2-8\left(x+y\right)+16\ge0\)

\(\Leftrightarrow\left(x+y-4\right)^2\ge0\)( ĐPCM )

Có : \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)

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

Theo BĐT Cô - si ta có :

\(\frac{x^2}{y-1}+4\left(y-1\right)\ge2\sqrt{\frac{x^2}{y-1}.4\left(y-1\right)}=4x\)

\(\frac{y^2}{x-1}+4\left(x-1\right)\ge4y\)

Do đó ; \(\frac{x^2}{y-1}+\frac{y^2}{x-1}+4.\left(x+y-2\right)\ge4\left(x+y\right)\)

\(\Leftrightarrow\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge8\)

Hay : \(P\ge8\)

Dấu "=" xảy ra khi \(x=y=2\)

Vậy \(P_{min}=8\) khi \(x=y=2\)

\(5x-\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\) 

\(5x-\left[4x-2\left(4x-15x+6\right)\right]=182\) 

\(5x-\left[4x-2\left(-11x+6\right)\right]=182\) 

\(5x-\left(4x+22x-12\right)=182\) 

\(5x-\left(26x-12\right)=182\) 

\(5x-26x+12=182\) 

\(-21x=182-12\) 

\(-21x=170\) 

\(x=-\frac{170}{21}\)