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1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(a.\left(2-3x\right)\left(x^2+2x+3\right)=0.\)
\(\left(2-3x\right)=0\)
\(\left(x^2+2x+3\right)=0\)
\(TH1:2-3x=0\Leftrightarrow x=\frac{-2}{-3}\)
\(TH2:x^2+2x+3=0\Leftrightarrow\left(x^2+2x+1\right)+3\Leftrightarrow\left(x+1\right)^2+3>0\)
b) \(3x-3x=5+2\) ( vô nghiệm)
c) vô nghiệm
d-\(x^2-5x-6=0\Leftrightarrow\left(x^2-x\right)+\left(6x-6\right)\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
vậy ...
x=1
x=-6
E) \(\frac{2\left(x-3\right)^2}{3}=\frac{3x^2}{2}\) quy đồng khử mẫu ta được
\(4\left(x-3\right)^2-9x^2=0\Leftrightarrow4\left(x-3\right)^2-\frac{4.1.9x^2}{4}\) rút 4 ta được
\(4\left\{\left(x-3\right)^2-\frac{9x^2}{4}\right\}=0\Leftrightarrow4\left\{\left(x-3\right)^2-\left(\frac{3}{2}x\right)^2\right\}\Leftrightarrow4\left(x-3+\frac{3}{2}x\right)\left(x-3-\frac{3}{2}x\right)=0\) ( hằng đẳng thức số 3 )
tích = 0
vậy ....
F) trị tuyệt đối + bình phương của 1 số thực luôn lớn hơn hoặc = 0( định lí Pain)
phá trị tuyệt đối ta được
\(\left(x+5\right)^2-\left(3x-2\right)^2=0\)
\(\left(x+5-3x-2\right)\left(x+5+3x-2\right)=0\) ( hẳng đẳng thức số 3 )
tích = 0 suy ra 2 TH vậy .....
g) câu G bạn lên coccoc math bạn ghi là nó ra kết quả phân tích thành nhân tử chứ làm = tay vừa dài vừa hại não :)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24=0\)
\(x\left(x-5\right)x\left(x^2-5x+10\right)=0\) ( coccoc math)
\(\left(x^2-5x+10\right)=0\Leftrightarrow\left(x^2-\frac{2x.5}{2}+\left(\frac{5}{2}\right)^2\right)+10-\frac{25}{4}=0\) ( 10-25/4) = 15/4
\(\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\) ( vô nghiệm)
vậy....