Cho biểu thức : A = \(\frac{2}{2x+3}\)+ \(\frac{3}{2x-3}\)- \(\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
a Rút gọn
b Tìm x để A = -1
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\(A=x^2-12x+7=x^2-12x+36-29\)
\(=\left(x-6\right)^2-29\ge-29\)
Vậy \(A_{min}=-29\Leftrightarrow x=6\)
\(C=x-x^2-4=-\left(x^2-x+4\right)\)
\(=-\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\)
\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{3}{4}\le-\frac{3}{4}\)
Vậy \(C_{min}=\frac{-3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(\left(a+4\right)^2-16a^2\)
\(=\left(a+4\right)^2-\left(4a\right)^2\)
\(=\left(a+4+4a\right)\left(a+4-4a\right)\)
\(=\left(5a+4\right)\left(4-3a\right)\)
\(x^3=4x\)
\(\Leftrightarrow x^3-4x=0\)
\(\Leftrightarrow x\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{4}=\pm2\end{cases}}\)
\(a,3x^3-6x^2+3x\)
\(=3x\left(x^2-2x+1\right)\)
\(=3x\left(x-1\right)^2\)
\(b,16x^2y-4xy^2-4x^3\)
\(=-4x\left(x^2-4xy+4y^2-3y^2\right)\)
\(=-4x\left(x-2y+y\sqrt{3}\right)\left(x-2y-y\sqrt{3}\right)\)
\(A=\left(x-1\right)^4-x^2\left(x^2+6\right)+4x\left(x^2+1\right)\)
\(A=x^4-4x^3+6x^2-4x+1-x^4-6x^2+4x^3+4x\)
\(A=\left(x^4-x^4\right)+\left(-4x^3+4x^3\right)+\left(6x^2-6x^2\right)+\left(-4x+4x\right)+1\)
\(A=1\)
Vậy biểu thức không phụ thuộc vào x
\(3x^3-14x^2+4x+3\)
\(=\left(3x^3-15x^2+9x\right)+\left(x^2-5x+3\right)\)
\(=3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)\)
\(=\left(3x+1\right)\left(x^2-5x+3\right)\)
3x^3 - 14x^2 + 4x + 3
= (3x^3+x^2) - 15^2- 5x+ 9x+ 3
= x^2(3x+1)- 5x(3x+1)+ 3(3x+1)
= (x^2- 5x+ 3)(3x+1)
#)Giải :
\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=\left(x+1\right)^4+x^2\left(x+1\right)^2+2x\left(x+1\right)+1\)
\(=\left(x+1\right)^2\left(2x^2+2x+1\right)+\left(2x^2+2x+1\right)\)
\(=\left(2x^2+2x+1\right)\left(x^2+2x+2\right)\)
\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=[\left(x+1\right)^2-x^2-x-1]\left[\left(x+1\right)^2+x^2+x+1\right]\)
\(=(x^2+2x+1-x^2-x-1)(x^2+x+1+x^2+x+1)\)
\(=x\left(2x^2+2x+2\right)\)
\(=2x\left(x^2+x+1\right)\)