1)\(x^2+6x+13=x^2+6x+9+4=\left(x+3\right)^2+4\)

Do \(\left(x+3\right)^2\ge0\)với mọi x

Nên \(\left(x+3\right)^2+4\ge4>0\)với mọi x 

Hay \(x^2+6x+13>0\)với mọi x

2/ Ta có: x²  + 6x + 13 = x² + 2.3x + 9 +4 = ( x + 3)² + 4

Ta có: (x+3)² >0 (với mọi x)

Nên (x+3)² + 4 \(\ge\)4 >0.

3/ Ta có: - x²+6x-11 = - (x²-6x+11)  = - (x²-2.3x+9+2) = - (x-3)²-2

Ta có: (x-3)²>0 với mọi x

Nên - (x-3)²<0 với mọi x

Suy ra - (x-3)²-2 \(\le\)- 2 <0

4/ Ta có: x -  y = 5 

Suy ra (x - y)² = 25

\(\Leftrightarrow\)  x² - 2xy + y²  = 25

\(\Leftrightarrow\)x² - 2.24  + y² = 25

\(\Leftrightarrow\)  x² + y² = 73

Ta có: x³ - y³ = (x - y).(x²  + xy + y² ) = 5.(73 + 24) =485

Bài 1 câu g bạn kia làm sai mình sửa lại nhá

\(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2\right)-12c^2\)

\(=3\left(a-b\right)^2-12c^2\)

\(=3\left[\left(a-b\right)^2-4c^2\right]\)

\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)

Để mình làm tiếp cho :))

Bài 2 :

Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)

\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)

\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)

\(=37,5.10-7,5.10\)

\(=10.30=300\)

Câu b : \(35^2+40^2-25^2+80.35\)

\(=\left(35^2+80.35+40^2\right)-25^2\)

\(=\left(30+45\right)^2-25^2\)

\(=75^2-25^2\)

\(=\left(75+25\right)\left(75-25\right)\)

\(=100.50=5000\)

Bài 3 :

Câu a : \(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)

Câu b : \(2x-2y-x^2+2xy-y^2=0\)

\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)

Câu c :

\(x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(x^2\left(x-3\right)+27-9x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)

Bài 4 :

Câu a :

\(x^2-4x+3\)

\(=x^2-x-3x+3\)

\(=\left(x^2-x\right)-\left(3x-3\right)\)

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(x-3\right)\)

Câu b :

\(x^2+x-6\)

\(=x^2-2x+3x-6\)

\(=x\left(x-2\right)+3\left(x-2\right)\)

\(=\left(x-2\right)\left(x+3\right)\)

Câu c :

\(x^2-5x+6\)

\(=x^2-2x-3x+6\)

\(=\left(x^2-2x\right)-\left(3x-6\right)\)

\(=x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(x-2\right)\left(x-3\right)\)

Câu d :

\(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

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

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

Bài 1:

a) \(2x^2-2xy-5x+5y\)

\(=\left(2x^2-2xy\right)-\left(5x-5y\right)\)

\(=2x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(2x-5\right)\)

b) \(8x^2+4xy-2ax-ay\)

\(=\left(8x^2+4xy\right)-\left(2ax+ay\right)\)

\(=4x\left(2x+y\right)-a\left(2x+y\right)\)

\(=\left(2x+y\right)\left(4x-a\right)\)

c) \(x^3-4x^2+4x\)

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

\(=x\left(x-2\right)^2\)

d) \(2xy-x^2-y^2+16\)

\(=-\left[\left(x^2-2xy+y^2\right)-16\right]\)

\(=-\left[\left(x-y\right)^2-4^2\right]\)

\(=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)

e) \(x^2-y^2-2yz-z^2\)

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

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

\(=-\left[\left(z+y+x\right)\left(z+y-x\right)\right]\)

g) \(3a^2-6ab+3b^2-12c^2\)

\(=\left(3a^2-6ab+3b^2\right)-12c^2\)

\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)

\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)

1, xy(x+y)+yz(y+z)+xz(x+z)+2xyz

= x²y+xy²+y²z+yz²+x²z+xz²+2xyz

=(x²y+x²z+xz²+xyz) + ( xy²+y²z+yz²+xyz)

=x(xy+xz+z²+yz)+y(xy+yz+z²+xz)

=(xy+xz+yz+z²).(x+y)

=(x(y+z)+z(y+z)).(x+y)

=((y+z).(x+z)).(x+y)= (x+y)(x+z)(y+z)

2. 3(x-3)(x-7)+(x-4)²+48

=3(x²+4x-21)+x²-8x+16+48

=4x²-4x+1 = (2x-1)²

Thay x=0,5 vào bt trên, ta có : (2.0,5 -1)²=0

3, x²-6x+10

= x²-2.3.x+9+1

=(x-3)²+1 \(\ge\)1 >0 ( do (x-3)² >=0 với mọi x)

=> x²6x+10 >0 với mọi x

4x-x²-5

=-(x²-4x+5)

=- (x²-2.2x+4+1)

= - ((x-2)²+1) = -(x-2)²-1\(\le\)-1 < 0 ( do (x-2)²\(\ge\)0 với mọi x => - (x-2)²\(\le\)0 với mọi x)

vậy, 4x-x²-5<0 với mọi x

Ta có : x² - 6x + 10 

= x² - 6x + 9 + 1 

= (x - 3)² + 1

Mà (x - 3)² \(\ge0\forall x\)

Nên : (x - 3)² + 1 \(\ge1\forall x\)

=> (x - 3)² + 1 \(>0\)(đpcm)

a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-11;1\right)\right\}\)

a: 

b: 

c: 

a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)

Bài 1: 

a,  \(x^2\) +2\(x\) = 0

     \(x.\left(x+2\right)\) = 0

     \(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

      \(x\) \(\in\) {-2; 0}

b, (-2.\(x\)).(-4\(x\)) + 28  = 100

      8\(x^2\)           + 28  = 100

        8\(x^2\)                   = 100 - 28

        8\(x^2\)                   = 72

          \(x^2\)                  = 72 : 8

          \(x^2\)                   = 9

           \(x^2\)                  = 3²

          |\(x\)|                  = 3

          \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\) 

Vậy \(\in\) {-3; 3}

c, 5.\(x\) (-\(x^2\)) + 1 = 6

   - 5.\(x^3\)       + 1 = 6

   5\(x^3\)                 = 1 - 6

   5\(x^3\)                 = - 5

    \(x^3\)                  =  -1

    \(x\)                    =  - 1

d, 3\(x^2\) + 12\(x\) = 0

   3\(x.\left(x+4\right)\) = 0

   \(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-4; 0}

e, 4.\(x.3\) = 4.\(x\)

    12\(x\) - 4\(x\) = 0

      8\(x\)          = 0

         \(x\)        = 0

a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x³ + y³ = 26

a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)

vậy............

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

\(\Rightarrow x^2+2xy+y^2=a^2\)

\(\Rightarrow xy=\frac{a^2-b}{2}\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)

Vậy....