xét dấu các biểu thức sau
a)f(x)=\(\frac{-4}{3x+1}-\frac{3}{2-x}\)
b)f(x)=4x2_1
c)f(x)=\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\)
Câu 1 : Xét dấu các biểu thức sau :
a , f(x) = \(\left(2x-1\right)\left(x+3\right)\)
b , f(x)= \(\left(-3x-3\right)\left(x+2\right)\left(x+3\right)\)
c , f(x) = \(\frac{-4}{3x+1}-\frac{3}{2-x}\)
d , f (x) = \(4x^2-1\)
e , f(x)= \(\left(-2x+3\right)\left(x-2\right)\left(x+4\right)\)
f , f(x) = \(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\)
g , f (x) = \(\frac{3}{2x-1}-\frac{1}{x-2}\)
h , f ( x) = \(\left(4x-1\right)\left(x+2\right)\left(3x-5\right)\left(-2x+7\right)\)
giúp mình với mình đang cần gấp
Bài 1. Xét dấu các biểu thức sau:
1. \(f\left(x\right)=\left(x-2\right)\left(5-3x\right)\left(x^2-x+3\right)\left(x^2+2x+1\right)\left(x^2-5x+4\right)\)
2. \(g\left(x\right)=\frac{5}{1-x}+\frac{5x}{x+1}+\frac{1}{x^2-1}\)
Bài 4 Xét dấu biểu thức sau
1 , \(f\left(x\right)=x^2-3x-2-\frac{8}{x^2-3x}\)
2 , \(f\left(x\right)=\frac{1}{x+1}-\frac{1}{x}-\frac{1}{2}\)
3 , \(f\left(x\right)=\frac{x^2-4x+3}{3-2x}-1+x\)
4 , \(f\left(x\right)=\frac{x^2-1}{\left(x^2-3\right)\left(-3x^2+2x+8\right)}\)
5 , \(f\left(x\right)=x^4-5x^2+2x+3\)
6 , \(f\left(x\right)=\frac{x^2+4x+15}{x^2-1}-\frac{x-3}{x+1}-\frac{x-2}{1-x}\)
1.
\(f\left(x\right)=\frac{\left(x^2-3x\right)^2-2\left(x^2-3x\right)-8}{x^2-3x}=\frac{\left(x^2-3x-4\right)\left(x^2-3x+2\right)}{x^2-3x}\)
\(f\left(x\right)=\frac{\left(x+1\right)\left(x-1\right)\left(x-2\right)\left(x-4\right)}{x\left(x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{0;3\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{-1;1;2;4\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -1\\0< x< 1\\2< x< 3\\x>4\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-1< x< 0\\1< x< 2\\3< x< 4\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{2x-2\left(x+1\right)-x\left(x+1\right)}{2x\left(x+1\right)}=\frac{-x^2-x-2}{2x\left(x+1\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{-1;0\right\}\)
\(f\left(x\right)>0\Rightarrow-1< x< 0\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\)
3.
\(f\left(x\right)=\frac{x^2-4x+3+\left(x-1\right)\left(3-2x\right)}{3-2x}=\frac{-x^2+x}{3-2x}=\frac{x\left(1-x\right)}{3-2x}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\frac{3}{2}\)
\(f\left(x\right)=0\Rightarrow x=\left\{0;1\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}0< x< 1\\x>\frac{3}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< 0\\1< x< \frac{3}{2}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(2-x\right)\left(3x+4\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\pm\sqrt{3};-\frac{4}{3};2\right\}\)
\(f\left(x\right)=0\Rightarrow x=\pm1\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}-\sqrt{3}< x< -\frac{4}{3}\\-1< x< 1\\\sqrt{3}< x< 2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -\sqrt{3}\\-\frac{4}{3}< x< -1\\1< x< \sqrt{3}\\x>2\end{matrix}\right.\)
5.
\(f\left(x\right)=x^4-x^3-x^2+x^3-x^2-x-3x^2+3x+3\)
\(=x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)-3\left(x^2-x-1\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-1\right)\)
Vậy:
\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=\frac{-1\pm\sqrt{13}}{2}\\x=\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{13}}{2}\\\frac{1-\sqrt{5}}{2}< x< \frac{1+\sqrt{5}}{2}\\x>\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{13}}{2}< x< \frac{1-\sqrt{5}}{2}\\\frac{1+\sqrt{5}}{2}< x< \frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{x^2+4x+15-\left(x-3\right)\left(x-1\right)+\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+7x+10}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x+5\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\pm1\)
\(f\left(x\right)=0\Rightarrow x=\left\{-2;-5\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\-2< x< -1\\x>1\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-5< x< -2\\-1< x< 1\end{matrix}\right.\)
Câu 1: Xét dấu các biểu thức sau:
a, f(x)=\(\frac{x-1}{\left(x+3\right)\left(2x-1\right)}\)
b, f(x)=(3x-1)(x2-4)
c, f(x)=\(\frac{\left(3x-1\right)\left(5-2x\right)}{\left(x+2\right)\left(x+5\right)}\)
Câu 2: Giải các bất phương trình:
a, (x-1)(x-2) ≥ 0
b, (3-x)(x+5) ≤ 0
c, \(\frac{2}{4-x}\) ≤ 1
d, \(\frac{4-3x}{x-2}\) ≥ 3
@Akai Haruma cô giúp em với ạ
Bài 3 : Xét dấu biểu thức sau :
1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\)
2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\)
3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\)
4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\)
5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\)
6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\)
7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\)
8 , \(f\left(x\right)=\left(-x^2+x-1\right)\left(6x^2-5x+1\right)\)
9 , \(f\left(x\right)=\frac{x^2-x-2}{-x^2+3x+4}\)
10 , \(f\left(x\right)=\left(x^2-5x+4\right)\left(2-5x+2x^2\right)\)
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:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
5.
\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)
\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
bài 1: xét dấu các nhị thức sau:
a) f(x)=3x+4
b) g(x)=4x-5
c) h(x)=-5x+10
d) k(x)=-6x-18
e) l(x)=2018-2x
bài 2: xét dấu các biểu thức sau:
a) f(x) =(2x-4)(3x+5)
b) g(x)=(-2x+8)(x-9)
c) h(x)=(-\(\frac{x}{3}\)-2)(1-2x)
d) k(x)=(1-2x)(x+1)(x-1)
e) l(x)=\(\frac{x-1}{x+1}\)
f) m(x)=\(\frac{4-2x}{3+x}\)
g) n(x)=\(\frac{\left(1-2x\right)\left(2x-1\right)}{3+x}\)
h) p(x)=\(\frac{x\left(x-1\right)\left(x+2\right)}{2-3x}\)
i) q(x)=\(\frac{\left(x-1\right)\left(2x-7\right)}{\left(2+x\right)^2}\)
Xét dấu các biểu thức tích, thương các tam thức bậc hai
a. \(f\left(x\right)=x^2\left(2-x-x^2\right)\left(x+2\right)\)
b. \(f\left(x\right)=\frac{x^4-3x^3+2x^2}{x^2-x-30}\)
a) F(x) = \(-x^2\left(x-1\right)\left(x+2\right)\left(x+2\right)=\left(1-x\right)x^2\left(x+2\right)^2\\ \)
\(\left\{{}\begin{matrix}x^2\ge0\\\left(x+2\right)^2\ge0\end{matrix}\right.\) => dấu biểu thức chỉ phụ thuộc vào thừa số (1-x)
F(x) =0 khi x={-2,0,1}
F(x) > 0 khi x<1 và khác -2 và 0
f(x) <0 khi x> 1
Tử f(x) =x^2(x^2-3x+2) =x^2(x-1)(x-2)
tương tự a) dấu của tử phụ thuộc (x-1)(x-2)
Mẫu f(x) =x^2 -x-30 =(x-5)(x+6)
Phần hỗ trợ Lập bảng đây khó thao tác
=> viết bằng hệ {điểm tới hạn xet x={-6,0,1,2,5}
Khi => \(\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)=>f(x) =0
Khi \(\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\) => f(x) không xác định
Khi \(x< -6\Rightarrow\left\{{}\begin{matrix}Tf\left(x\right)>0\\Mf\left(x\right)>0\end{matrix}\right.\)\(\Rightarrow f\left(x\right)>0\)
khi -6<x<1 \(\Rightarrow\left\{{}\begin{matrix}Tf\left(x\right)>0\\Mf\left(x\right)< 0\end{matrix}\right.\) => f(x) <0
khi 1<x<2 \(\Rightarrow\left\{{}\begin{matrix}Tf\left(x\right)< 0\\Mf\left(x\right)< 0\end{matrix}\right.\) => f(x) >0
khi 2<x<5 \(\Rightarrow\left\{{}\begin{matrix}Tf\left(x\right)>0\\Mf\left(x\right)< 0\end{matrix}\right.\) => f(x) <0
khi x>5 \(\Rightarrow\left\{{}\begin{matrix}Tf\left(x\right)>0\\Mf\left(x\right)>0\end{matrix}\right.\) => f(x) >0
Bài 2 Xét dấu biểu thức sau
1 , \(f\left(x\right)=x^2-\sqrt{3}x+\frac{3}{4}\)
2 , \(f\left(x\right)=-x^2+3x-2\)
3 , \(f\left(x\right)=x^4-4x+1\)
4 , \(f\left(x\right)=\frac{3x+7}{x^2-x-2}\)
5 , \(f\left(x\right)=\frac{x+2}{3x+1}-\frac{x-2}{2x-1}\)
6 , \(f\left(x\right)=\frac{1}{x^2-5x+4}-\frac{1}{x^2-7x+10}\)
7 , \(f\left(x\right)=\left(x-1\right)\left(x-3\right)-\frac{18}{x^2-4x-4}\)
8 , \(f\left(x\right)=\left(x^2-1\right)\left(x-2\right)\)
9 , \(f\left(x\right)=\left(x+3\right)\left(-4x^2+9x-2\right)\)
10 , \(f\left(x\right)=\frac{10-x}{5+x^2}-\frac{1}{2}\)
Bài 1: xét dấu các biểu thức sau: (giải chi tiết, cặn kẽ giúp mình nhe)
a)\(f\left(x\right)\frac{1}{3-x}-\frac{1}{3+x}\)
b)\(f\left(x\right)=2x^2+2x+5\)
c)\(f\left(x\right)=\left(2x-1\right)\left(x^2-4\right)\)
d)\(f\left(x\right)=\frac{4x^2-19x+12}{x-7}\)