\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-2\right)+c\left(a+b\right)^2\left(a-b\right)\)

\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)

nguồn câu hỏi tương tự

Trang 136 trong nâng cao phát triển có viết rồi mình cóp nó vô để mọi người dễ đọc nhé !

\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)

\(=\left[a.\left(a+b+c\right)+bc\right]\left[b.\left(a+b+c\right)+ac\right]\left[c.\left(a+b+c\right)+ab\right]\)

\(=\left(a^2+ab+ac+bc\right)\left(ba+b^2+bc+ac\right)\left(ca+cb+c^2+ab\right)\)

\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ba+b^2\right)+\left(bc+ac\right)\right]\left[\left(ca+c^2\right)\left(cb+ab\right)\right]\)

\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(b+a\right)\right]\left[c\left(a+c\right)b\left(b+b\right)\right]\)

\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)

\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

\(\Rightarrowđpcm\)

\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)

\(=\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ac\right]\left[c\left(a+b+c\right)+ab\right]\)

\(=\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)\left(ac+bc+c^2+ab\right)\)

\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ab+b^2\right)+\left(bc+ac\right)\right]\left[\left(ac+c^2\right)+\left(bc+ab\right)\right]\)

\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(a+c\right)+b\left(a+c\right)\right]\)

\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

\(\Rightarrowđpcm\)

\(4x^4-10x^3+8x^2-5x-1=0\)

\(\left(x^4-x^3+2x^2\right)-\left(4x^3-4x^2+8x\right)+\left(2x^2-2x+4\right)=0\)

\(x^2\left(x^2-x+2\right)-4x\left(x^2-x+2\right)+2\left(x^2-x+2\right)=0\)

\(\left(x^2-x+2\right)\left(x^2-4x+2\right)=0\)

\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]\left(x^2-4x+2\right)=0\)

Vì \(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)\(\Rightarrow x^2-4x+2=0\)

\(\Rightarrow\left(x-2\right)^2=2\)\(\Rightarrow x-2=\pm\sqrt{2}\)

\(\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+2\\x=2-\sqrt{2}\end{cases}}\)

( x - 5 )⁴ + ( x - 3 )⁴ = 16

Đặt x - 4 = a

\(\Rightarrow\)x - 5 = a -1 

        x - 3 = a +1

Khi đó phương trình trở thành:

 ( a  - 1 )⁴ + ( a + 1 )⁴ = 16

\(\Leftrightarrow\)[ ( a - 1 )⁴  + 2.( a - 1 )².( a + 1 )² + ( a + 1 )⁴  ] - 2.( a - 1 )².( a + 1 )² = 16

\(\Leftrightarrow\)[ ( a - 1 )² + ( a + 1 )² ]² - 2.( a - 1 )².( a + 1 )² = 16

\(\Leftrightarrow\)( a² - 2a + 1 + a² + 2a + 1 )² - 2.( a² - 1 )² = 16

\(\Leftrightarrow\)( 2a² + 2 )²  - 2.( a⁴ - 2a² + 1 ) = 16

\(\Leftrightarrow\)4a⁴ + 8a² + 4 - 2a⁴ + 4a² - 2 - 16 = 0

\(\Leftrightarrow\) 2a⁴ + 12a² - 14 = 0

\(\Leftrightarrow\)2.( a⁴ + 6a² - 7 ) = 0

\(\Leftrightarrow\) a⁴ + 6a² - 7 = 0

\(\Leftrightarrow\) a⁴ + 7a² - a² - 7 = 0

\(\Leftrightarrow\) a².( a² + 7 ) - ( a² + 7 ) = 0

\(\Leftrightarrow\)( a² - 1 ).( a² + 7 ) = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}a^2-1=0\\a^2+7=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a=\pm1\\a^2=-7\left(lo\text{ại}\right)\end{cases}}\)

Với a = 1                                              Với a = -1

\(\Rightarrow\) x - 4 = 1                                     \(\Rightarrow\) x - 4 = -1

\(\Leftrightarrow\) x = 5                                          \(\Leftrightarrow\) x = 3

Vậy x = 5 , x = 3

\(4x^2+y^2-2x-y-2xy+1=1\) 

\(\Leftrightarrow4x^2-4xy+y^2-2x-y+2xy=0\) 

\(\Leftrightarrow\left(2x-y\right)^2-2x-y+2xy=0\) 

\(\Leftrightarrow x\left[\left(2x-y\right)-2x-y+2xy\right]=0\) 

\(\Leftrightarrow x\left(2x-y\right)^2-2x^2+xy=0\) 

\(\Leftrightarrow x\left[\left(2x-y\right)^2-2x+y\right]=0\) 

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(2x-y\right)^2-2x+y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(2.0-y\right)^2-2.0+y=0\end{cases}}}\) (thay x=0 vào biểu thức dưới)

\(\Leftrightarrow x=0\) hoặc  \(y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\\y=-1\end{cases}}\)  (mà x;y nguyên dương )=>y=0

Vậy x=0 ;y=0

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\left(tm\right)\\y=-1\left(ktm\right)\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\left(tm\right)\\y=-1\left(ktm\right)\end{cases}}\end{cases}}\)

Bạn sai rồi nhé. Khi ta giải đc x=0 ở Th1 thì không được áp dụng x=0 ở th2

\(3x\left(x+7\right)+21-3x^2=0\)

\(\Leftrightarrow3x^2+21x+21-3x^2=0\)

\(\Leftrightarrow21x+21=0\)

\(\Leftrightarrow21\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(5x\left(3x+7\right)-15x^2=70\)

\(\Leftrightarrow15x^2+35x-15x^2=70\)

\(\Leftrightarrow35x=70\)

\(\Leftrightarrow x=2\)

#)Giải :

Ta có : \(a^4+b^4+c^4+d^4=4abcd\)

\(\Leftrightarrow a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2c^2d^2=0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-d^2\right)+2\left(ab-cd\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a^2=b^2\\c^2=d^2\\ab=cd\end{cases}}\)

Do a, b, c, d > 0

\(\Leftrightarrow a=b=c=d\left(đpcm\right)\)