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a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
\(\text{*Với }x-3\ge0\text{ thì:}\)
\(A=5\left(x-3\right)-2\left(2x-1\right)\)
\(=5x-15-4x+2\)
\(=x-13\)
\(\text{*Với }x-3< 0\text{ thì:}\)
\(A=-5\left(x-3\right)-2\left(2x-1\right)\)
\(=-5x+15-4x+2\)
\(=-9x+17\)
\(\cdot\text{Vậy:}\)
\(A=x-13\text{ khi }x-3\ge0\)
\(A=-9x+17\text{ khi }x-3< 0\)
Gọi biểu thức trên là T
+)Xét \(x-3\ge0\Leftrightarrow x\ge3\)
T trở thành:\(T=3\left(x-1\right)-2\left(x-3\right)\)
\(=\left(3x-2x\right)-\left(3-6\right)\)\(=x+3\) (1)
+)Xét \(x-3< 0\Leftrightarrow x< 3\)
Khi đó: \(T=3\left(x-1\right)-2\left[-\left(x-3\right)\right]\)
\(=3\left(x-1\right)-2\left(-x+3\right)\)
\(=\left(3x+2x\right)-\left(3+6\right)=5x-9\)(2)
Từ (1) và (2) ...
\(B=2\left|x+1\right|-\left|x-1\right|\)
- Nếu \(x\ge1\) thì : \(B=2.\left(x+1\right)-\left(x-1\right)=2x+2-x+1=x+3\)
- Nếu \(x=0\) thì :
\(B=2.\left(0+1\right)-\left(0-1\right)=2-\left(-1\right)=3\)
- Nếu \(x< 0\) thì :
\(B=2.\left(-x-1\right)-\left(-x+1\right)=-2x-2+x-1=-x-3\)
Nếu x < 1 => |-x + 1| = -x + 1
|2x - 3| = -(2x - 3) = -2x + 3
Khi đó B = |-x + 1| + |2x - 3| - 2(x - 1)
= -x + 1 - 2x + 3 - 2x + 2
= - 5x + 6
Nếu \(1\le x\le1,5\)
=> |-x + 1| = x - 1
|2x - 3| = --2x + 3
Khi đó B = x - 1 - 2x + 3 - 2x + 2
= -4x + 4
Nếu x > 1,5 => |-x + 1| = x - 1
|2x - 3| = 2x - 3
Khi đó B = x - 1 + 2x - 3 -2x + 2
= x - 2
how to đăng bài?
please help me