\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)

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

\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)

\(=10x^2+16x+50\)

B1:

Vì \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y-\frac{1}{3}\right|\ge0\\\left|4z+5\right|\ge0\end{cases}\left(\forall x,y,z\right)}\Rightarrow\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\ge0\left(\forall x,y,z\right)\)

Mà theo đề bài, \(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\le0\) nên dấu "=" xảy ra khi:

\(\left|x-\frac{1}{2}\right|=\left|2y-\frac{1}{3}\right|=\left|4z+5\right|=0\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{6}\\z=-\frac{5}{4}\end{cases}}\)

B2:

a) Nếu \(x< 1\) => \(A=1-x+x+3=4\)

Nếu \(x\ge1\) => \(A=x-1+x+3=2x+2\)

b) Nếu \(x< -\frac{3}{2}\) => \(B=2x+2x+3=4x+3\)

Nếu \(x\ge-\frac{3}{2}\) => \(B=2x-2x-3=-3\)

Bài 1.

Ta có \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\forall x\\\left|2y-\frac{1}{3}\right|\ge0\forall y\\\left|4z+5\right|\ge0\forall z\end{cases}}\Rightarrow\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\ge0\forall x,y,z\)

Kết hợp với đề bài => Chỉ xảy ra trường hợp \(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\2y-\frac{1}{3}=0\\4z+5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{6}\\z=-\frac{5}{4}\end{cases}}\)

Bài 2.

A = | x - 1 | + x + 3

Với x < 1 => A = -( x - 1 ) + x + 3 = -x + 1 + x + 3 = 4

Với x ≥ 1 => A = ( x - 1 ) + x + 3 = x - 1 + x + 3 = 2x + 2

B = 2x - | 2x + 3 |

Với x < -3/2 => B = 2x - -( 2x + 3 ) = 2x + ( 2x + 3 ) = 2x + 2x + 3 = 4x + 3 

Với x ≥ -3/2 => B = 2x + -( 2x + 3 ) = 2x - ( 2x + 3 ) = 2x - 2x - 3 = -3

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

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

\(=10x^2+40x+50\)

\(=\left(x^2+10x+25\right)+\left(9x^2+30x+25\right)\)

\(=\left(x+5\right)^2+\left(3x+5\right)^2\)

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

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

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

\(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)\)

\(=\left(3x+2\right)^2+\left(x+5^2\right)\)

Bài 2 :

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

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

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

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

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

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

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

10x²+40x+50

Ta có :

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

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

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

\(=\left(x^2+10x+25\right)+\left(9x^2+30x+25\right)\)

\(=\left(x+5\right)^2+\left(3x+5\right)^2\)

Vậy biểu thức trên viết được dưới dạng tổng các bình phương của 2 biểu thức(đpcm)

Giải:

a) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)

\(=\left[\left(x^2+x-1\right)-\left(x^2+2x+3\right)\right]\left[\left(x^2+x-1\right)+\left(x^2+2x+3\right)\right]\)

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

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

Vậy ...

b) \(-16+\left(x-3\right)^2\)

\(=\left(x-3\right)^2-16\)

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

\(=\left(x-3-4\right)\left(x-3+4\right)\)

\(=\left(x-7\right)\left(x+1\right)\)

Vậy ...

c) \(64+16y+y^2\)

\(=8^2+2.8.y+y^2\)

\(=\left(8+y\right)^2\)

Vậy ...

a. \(9x^2+30x+25=\left(3x+5\right)^2\)

b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)

c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)

d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)

e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)