`Answer:`

`(x+5)(2x-1)=0`

`<=>x+5=0` hoặc `2x-1=0`

`<=>x=-5` hoặc `2x=1`

`<=>x=-5` hoặc `x=1/2`

`Answer:`

`a)7x+21=0`

`<=>7x=-21`

`<=>x=-21:7`

`<=>x=-3`

`b)3x+1=7x-11`

`<=>3x-7x=-11-1`

`<=>-4x=-12`

`<=>x=3`

`c)4/3x-5/6=1/2`

`<=>4/3x=1/2+5/6`

`<=>4/3x=4/3`

`<=>x=1`

`d)\frac{x-3}{5}=1-\frac{1-2x}{3}`

`<=>3(x-3)=15-5(1-2x)`

`<=>3(x-3)-15+5(1-2x)=0`

`<=>3x-9-15+5-10x=0`

`<=>-7x-19=0`

`<=>-7=19`

`<=>x=\frac{-19}{7}`

`e)\frac{2x}{3}+\frac{2x-1}{6}=4-x/3`

`<=>2.2x+2x-1=4.6-2x`

`<=>4x+2x-1=24-2x`

`<=>6x+2x=24+1`

`<=>8x=25`

`<=>x=\frac{25}{8}`

`f)(4x-10)(24+5x)=0`

`<=>2(2x-5)(24+5x)=0`

`<=>2x-5=0` hoặc `24+5x=0`

`<=>x=5/2` hoặc `x=\frac{-24}{5}`

`g)x^2+1=x(x-1)`

`<=>x^2+1=x^2-x`

`<=>x^2-x^2=-x-1`

`<=>-x-1=0`

`<=>-x=1`

`<=>x=-1`

`h)

`i)\frac{2x^2-3x-2}{x^2-4}=2(ĐKXĐ:x\ne+-2)`

`<=>\frac{2x^2-4x+x-2}{(x-2)(x+2)}=2`

`<=>\frac{2x+1}{x+2}=2`

`<=>2x+1=2x+4`

`<=>1=4` (Vô lý)

Vậy phương trình vô nghiệm.

`i)\frac{x-1}{x+1}+3=\frac{2x+3}{x+1}(ĐK:x\ne-1)`

`<=>3=\frac{2x+3}{x+1}-\frac{x-1}{x+1}`

`<=>\frac{2x+3-x+1}{x+1}=3`

`<=>\frac{x+4}{x+1}=3`

`<=>3(x+1)=x+4`

`<=>3x+3=x+4`

`<=>2x=1`

`<=>x=1/2`

`Answer:`

Vế trái: `(x^2-6x+11)(y^2+2y+4)`

`=((x-3)^2+2)((y+1)^2+3)`

`>=2.3`

`>=6(1)`

Vế phải: `-z^2+4z+2`

`=-(z^2-4z-2)`

`=-(z^2-4z+4-6)`

`=-(z-2)^2+6=<6(2)`

Từ `(1)(2)` suy ra dấu "=" xảy ra khi: \(\hept{\begin{cases}x-3=0\\y+1=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\\z=2\end{cases}}}\)

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Ta có 

a²+b²+c² = ab+bc+ca

<=> 2(a²+b²+c²)= 2(ab+bc+ca)

<=> (a - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> a = b = c

Thế vào pt thứ (2) ta được

a⁸ + b⁸ + c⁸ = 3

<=> 3a⁸ = 3

<=> a⁸ = 1

<=> a = b = c = 1(3) hoặc a = b = c = - 1(4)

Từ (3) => P = 1 + 1 - 1 = 1

Từ (4) => P = - 1 + 1 + 1 = 1

ta có   :\(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2.\left(a^2+b^2+c^2\right)=2.\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

mà ta có:  \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)   \(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)  \(\forall a,b,c\)

dấu  \("="\) xảy ra \(\Leftrightarrow a=b=c\)

lại có:\(a^8+b^8+c^8=3\)  mà \(a=b=c\)

\(\Rightarrow a^8+a^8+a^8=3\)

\(\Leftrightarrow a^8=1\)

\(\Leftrightarrow a=1\)

vậy \(a=b=c=1\)

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A=\(x.\left(x+2\right).\left(x^2+2x+2\right)+1\)

\(=x.\left(x+2\right).\left[x.\left(x+2\right)+2\right]+1\)

đặt \(x.\left(x+2\right)=a\) ta có:

\(A=a.\left(a+2\right)+1\)

\(A=a^2+2a+1\)

\(=\left(a+1\right)^2\)

\(=\left[x.\left(x+2\right)+1\right]^2\)

\(=\left(x^2+2x+1\right)^2\)

\(=\left(x+1\right)^4\)

`Answer:`

`a^2.(b-c)+b^2.(c-a)+c^2.(a-b)`

`=a^2.(b-c)+b^2[(c-b)-(a-b)]+c^2.(a-b)`

`=a^2.(b-c)+b^2.(c-b)+b^2.(a-b)+c^2.(a-b)`

`=(b-c)(a^2-b^2)-(a-b)(b^2-c^2)`

`=(b-c)(a-b)(a+b)-(a-b)(b-c)(b+c)`

`=(a-b)(b-c)(a+b-b-c)`

`=(a-b)(b-c)(a-c)`

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

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

\(=a^2.\left(b-c\right)-b^2.\left[\left(a-b\right)+\left(b-c\right)\right]+c^2.\left(a-b\right)\)

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

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

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

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

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