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\(4-\left|2x+1\right|=3\)
\(\Leftrightarrow\left|2x+1\right|=4-3\)
\(\Leftrightarrow\left|2x+1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy tập ngiệm \(S=\left\{0;-1\right\}\)
11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)
12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
13)
\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)
14)
\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)
Hướng dẫn 1 câu, các câu khác làm tương tự:
\(A\left(x\right)=2x^3+7x^2+ax+b\)
\(=\left(2x^3+2x^2-2x\right)+\left(5x^2+5x-5\right)-3x+5+ax+b\)
\(=2x\left(x^2+x-1\right)+5\left(x^2+x-1\right)+x\left(a-3\right)+\left(b+5\right)\)
Để \(A\left(x\right)⋮B\left(x\right)\) thì:
\(\left\{{}\begin{matrix}a-3=0\\b+5=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)
b) Thực hiện phép chia \(A\left(x\right)\) cho \(B\left(x\right)\)
ax3+bx-24 l x2+4x+3
ax3+4ax2+3ax ax-4a
________________
-4ax2+x(b-3a)-24
-4ax2-16ax-12a
__________________
x(b+13a)+12a-24
Để \(A\left(x\right)⋮B\left(x\right)\) thì:
\(\left\{{}\begin{matrix}b+13a=0\\12a-24=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=-26\end{matrix}\right.\)
a: \(=\dfrac{x+1+1}{x+2}=1\)
b: \(=\dfrac{3x-7+3x+7}{6}=x\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x=t\)
\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)
\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)
Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)