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Rút gọn các biểu thức 
\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)
\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)
\(A=-12x\)
\(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)
\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)
\(B=-3x-12\)
Câu C tương tự.
Chúc bạn học tốt!!!
A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)
A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)
A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)
A = \(-4x^2-12x-8+4x^2+8=-12x\)
b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)
B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)
B = \(-6x^2-3x-6\)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
Bài 1:
a) \(9x^2-6x+2\)
\(\Leftrightarrow9x^2-6x+1+1\)
\(\Leftrightarrow\left(3x-1\right)^2+1\)
Vì \(\left(3x-1\right)^2\ge0\forall x,1>0\)
\(\Rightarrow9x^2-6x+2\) luôn dương với mọi x.
b) \(x^2+x+1\)
\(\Leftrightarrow x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x,\dfrac{3}{4}>0\)
\(\Rightarrow x^2+x+1\) luôn dương với mọi x.
Bài 2 :
a) \(A=x^2-3x+5\)
\(\Leftrightarrow A=x^2-3x+2+3\)
\(\Leftrightarrow A=\left(x-2\right)\left(x-1\right)+3\)
Vì \(\left(x-2\right)\left(x-1\right)\ge0\forall x\) => \(A\ge3\)
Vậy GTNN A đạt được = 3 khi và chỉ khi x = 2 hoặc x = 1.
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(\Leftrightarrow B=4x^2-4x+1+x^2+4x+4\)
\(\Leftrightarrow B=5x^2+5\)
\(\Leftrightarrow B=5\cdot\left(x^2+1\right)\)
Vì \(x^2+1\ge1\forall x\)
=> GTNN của B đạt được = 5 khi và chỉ khi x = 0.
Bài 3 :
a) \(A=-x^2+2x+4\)
Làm tương tự ta có \(A_{MAX}=5\) khi và chỉ khi x = 1.
b) \(B=-x^2+4x\)
Làm tương tự ta có \(B_{MAX}=4\) khi và chỉ khi x = 2.
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
\(a.M=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
\(vt=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=\left(x^3-3.x^2.1+3.x.1^2-1^3\right)-\left[4x.x+4x.1+4x.x+4x.\left(-1\right)+x.x+x.\left(-1\right)+1.x+1.\left(-1\right)\right]\)
\(=\left(x^3-3x^2+3x-1\right)-\left(4x^2+4x+4x^2-4x+x^2-x+x-1\right)\)
\(=\left(x^3-3x^2+3x-1\right)-\left(2.4x^2+x^2-1\right)\)
\(=x^3-3x^2+3x-1-8x^2-x^2+1\)
\(=x^3-12x^2\)
\(vp=3\left(x-1\right)\left(x^2+x+1\right)\)
\(=3\left(x^3-1^3\right)\)
\(=3x^3-3\)
\(\Rightarrow M=x^3-12x^2=3x^3-3\)
\(\Rightarrow M=x^3-12x^2-3x^3+3\)
\(\Rightarrow M=-2x^3-12x^2+3\)
\(b.khix=-2,tacó:\)
\(M=-2.\left(-2\right)^3-12.\left(-2\right)^2+3\)
\(=-2.\left(-8\right)-12.4+3\)
\(=16-48+3\)
\(=-29\)
Bài 1:
a: \(\Leftrightarrow x^2-4x-x^2+8=0\)
=>-4x+8=0
hay x=2
b: \(\Leftrightarrow3x^2-3x+2x-2-3\left(x^2-x-2\right)=4\)
\(\Leftrightarrow3x^2-x-2-3x^2+3x+6=4\)
=>2x+4=4
hay x=0
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow-3=5x\)
\(\Rightarrow5x=-3\)
\(\Rightarrow x=-\dfrac{3}{5}\)
Vậy ....
P/s : Làm bừa !
a/ \(M=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)
\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x-1}{x^2+1}\)
\(=\dfrac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(=\dfrac{2x^2+1}{x-1}\)
===========
b/ Thay x=-4 vào M ta được:
\(\dfrac{2.\left(-4\right)^2+1}{-4-1}=-\dfrac{33}{5}\)
Vậy: Giá trị của M tại x=-4 là \(-\dfrac{33}{5}\)