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

Ta có bảng xét dấu:

(*) Nếu \(x\le-4\), ta có:

PT (1):\(4\left(2-x\right)-3\left(-4-x\right)=3\)

\(\Leftrightarrow8-4x+12+3x=3\)

\(\Leftrightarrow20-x=3\)

\(\Leftrightarrow-x=-17\)

\(\Leftrightarrow x=17\)( không thỏa mãn )

(*) Nếu \(-4< x\le2\), ta có:

PT (1): \(4\left(2-x\right)-3\left(4+x\right)=3\)

\(\Leftrightarrow8-4x-12-3x=3\)

\(\Leftrightarrow-4-7x=3\)

\(\Leftrightarrow-7x=7\)

\(\Leftrightarrow x=-1\)( thỏa mãn )

(*) Nếu \(x>2\), ta có :

PT (1):\(4\left(2-x\right)-3\left(4+x\right)=3\)

\(\Leftrightarrow-8+4x-12-3x=3\)

\(\Leftrightarrow-20+x=3\)

\(\Leftrightarrow x=23\)( thỏa mãn )

Vậy phương trình có tập nghiệm \(S=\left\{23;-1\right\}\)

d) mk chỉnh lại đề

  \(8xy^2-5xyz-24y+15z\)

\(=xy\left(8y-5z\right)-3\left(8y-5z\right)\)

\(=\left(8y-5z\right)\left(xy-3\right)\)

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

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

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

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

i)  \(2x^3+x^2-4x-12=\left(x-2\right)\left(2x^2+5x+6\right)\)

k)  \(25x^2\left(x-5\right)-x+y=\left(1-5x\right)\left(1+5x\right)\left(y-x\right)\)

\(C=4x^2+3+4x\)

\(C=\left[\left(2x\right)^2+2.2x+1\right]+2\)

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

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2+2\ge2\forall x\)

\(C=2\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(C=2\Leftrightarrow x=-\frac{1}{2}\)

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

\(\left(x+3\right).\left(x+3\right)-xx-x=6\)

\(x\left(x+3\right)+3\left(x+3\right)-11x-x=6\)

\(x.x+3x+3x+9-\left(11x+x\right)=6\)

\(x.\left(3+3+x\right)-x\left(11+1\right)=6\)

\(x.\left(6+x\right)-x\left(11+1\right)=6\)

\(x.\left(6+x\right)-12x=6\)

\(x.\left(6+x-12\right)=6\)

\(x.\left(-6+x\right)=6\)

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

\(x\left(x-6\right)=6\)

\(6\left(x-6\right)=x\)

\(6x-36=x\)

\(6x-x=36\)

\(5x=36\)

\(x=\frac{36}{5}\)

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

\(x^2+6x+9-x^2+x=6\)

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

\(7x=-3\)

\(x=-\frac{3}{7}\)

\(\left(x-1\right)^2-\left(x+3\right)\left(x-3\right)=55\)

\(\Leftrightarrow\)\(x^2-2x+1-x^2+9=55\)

\(\Leftrightarrow\)\(-2x=45\)

\(\Leftrightarrow\)\(x=-\frac{45}{2}\)

Vậy...

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

Ta có bảng xét dấu:

(*) Nếu \(x< -2\)

PT (1)\(\Leftrightarrow2\left(1-2x\right)-3\left(-x-2\right)=3\)

\(\Leftrightarrow2-4x+3x+6=3\)

\(\Leftrightarrow8-x=3\)

\(\Leftrightarrow-x=-5\)

\(\Leftrightarrow x=5\)(không thỏa mãn )

(*) Nếu \(2\le x< \frac{1}{2}\)

PT (1) \(\Leftrightarrow2\left(1-2x\right)-3\left(x+2\right)=3\)

\(\Leftrightarrow2-4x-3x-6=3\)

\(\Leftrightarrow-7x-4=3\)

\(\Leftrightarrow x=-1\) ( thỏa mãn )

(*) Nếu \(x\ge\frac{1}{2}\)

PT (1) \(\Leftrightarrow2\left(2x-1\right)-3\left(x+2\right)=3\)

\(\Leftrightarrow4x-2-3x-6=3\)

\(\Leftrightarrow x-8=3\)

\(\Leftrightarrow x=11\) ( thỏa mãn )

Vậy phương trình có tập nghiệm \(S=\left\{11;-1\right\}\)

a)  \(\left(x+y\right)^2=\left(-7\right)^2=49\)

b)  \(x^2+y^2=\left(x+y\right)^2-2xy=49-2.12=25\)

c)   \(x^3+y^3=\left(x+y\right)\left(x^2+y^2\right)-xy\left(x+y\right)\)

\(=\left(-7\right).25-12\left(-7\right)=-91\)

d)  \(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=25^2-2.12^2=337\)

p/s: mấy câu còn lại lm tương tự nhé