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Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
Câu 1:
a: Sửa đề: \(A=\left(x+2\right)\left(x^2-2x+4\right)+x\left(1-x\right)\left(1+x\right)\)
\(=x^3+2^3+x\left(1-x^2\right)\)
\(=x^3+8+x-x^3\)
=x+8
b: Khi x=-4 thì A=-4+8=4
c: Đặt A=-2
=>x+8=-2
=>x=-10
Câu 2:
a: \(x^3-3x^2=x^2\cdot x-x^2\cdot3=x^2\left(x-3\right)\)
b: \(5x^3+10x^2+5x\)
\(=5x\cdot x^2+5x\cdot2x+5x\cdot1\)
\(=5x\left(x^2+2x+1\right)\)
\(=5x\left(x+1\right)^2\)
a,Rút gọn biểu thức sau :2x(x+3)-x(2x-1)
2x(x+3)-x(2x-1) = 2x^2 + 6x - 2x^2 + x = 7x
a: \(=2x^2+6x-2x^2+x=7x\)
b: \(\Leftrightarrow x^3+2x^2-x^2-2x+x+2+a-2⋮x+2\)
=>a-2=0
=>a=2
c: \(36x^2-49=0\)
=>(6x-7)(6x+7)=0
=>x=7/6 hoặc x=-7/6
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
=>x^3+27-x^3+x-27=0
=>x=0