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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) 3x^3 -10x+3 =(3x-1)(x-3)

Kết luận

VT< 0 {dấu "-"} khi x <1/3 hoắc 5/4<x<3

VT>0 {dấu "+"} khi x 1/3<5/4 hoặc x> 3

VT=0 {không có dấu} khi x={1/3;5/4;3}

a/

\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)

\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)

b/

\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)

\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)

c/

\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)

\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)

d/ \(x\ge\frac{3}{2}\)

\(\Leftrightarrow\left(5x+1\right)^2=\left(2x-3\right)^2\)

\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)

e/

\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)

f/

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)

g/

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x^2-x-2\\x^2-2x=-2x^2+x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)

a) Ta lập bảng xét dấu

Kết luận: f(x) < 0 nếu - 3 < x <

f(x) = 0 nếu x = - 3 hoặc x =

f(x) > 0 nếu x < - 3 hoặc x > .

b) Làm tương tự câu a).

f(x) < 0 nếu x ∈ (- 3; - 2) ∪ (- 1; +∞)

f(x) = 0 với x = - 3, - 2, - 1

f(x) > 0 với x ∈ (-∞; - 3) ∪ (- 2; - 1).

c) Ta có: f(x) =

Làm tương tự câu b).

f(x) không xác định nếu x = hoặc x = 2

f(x) < 0 với x ∈ ∪

f(x) > 0 với x ∈ ∪ (2; +∞).

d) f(x) = 4x² – 1 = (2x - 1)(2x + 1).

f(x) = 0 với x =

f(x) < 0 với x ∈

f(x) > 0 với x ∈ ∪

a/ \(x\ge-3\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)

b/ \(x\ge-\frac{5}{2}\)

\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\)

\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

c/ \(x\ge1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=5x-5\\2x^2-3x-5=5-5x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x=0\\2x^2+2x-10=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\\x=\frac{-1+\sqrt{21}}{2}\\x=\frac{-1-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)

d/ \(x\ge\frac{17}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\left(l\right)\\x=\sqrt{22}\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)

e/ \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-x-2=x-2\\3x^2-x-2=2-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=0\\3x^2=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{2}{3}\left(l\right)\\x=\frac{2\sqrt{3}}{3}\\x=\frac{-2\sqrt{3}}{3}\end{matrix}\right.\)