1:

c: =>1/3x+2/3-x+1>x+3

=>-2/3x+5/3-x-3>0

=>-5/3x-4/3>0

=>-5x-4>0

=>x<-4/5

d: =>3/2x+5/2-1<=1/3x+2/3+x

=>3/2x+3/2<=4/3x+2/3

=>1/6x<=2/3-3/2=-5/6

=>x<=-5

2:

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) 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}

\(2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\);\(x-1=0\Leftrightarrow x=1\); \(x+2=0\Leftrightarrow x=-2\).
TenAnh1 TenAnh1 B = (11.24, -6.26) B = (11.24, -6.26) B = (11.24, -6.26) C = (-0.38, -6.9) C = (-0.38, -6.9) C = (-0.38, -6.9) D = (14.98, -6.9) D = (14.98, -6.9) D = (14.98, -6.9) E = (-0.4, -6.68) E = (-0.4, -6.68) E = (-0.4, -6.68) F = (13.84, -6.58) F = (13.84, -6.58) F = (13.84, -6.58) G = (0.82, -8.58) G = (0.82, -8.58) G = (0.82, -8.58) H = (16.18, -8.58) H = (16.18, -8.58) H = (16.18, -8.58) I = (-0.56, -6.62) I = (-0.56, -6.62) I = (-0.56, -6.62) J = (14.8, -6.62) J = (14.8, -6.62) J = (14.8, -6.62) K = (-0.36, -6.84) K = (-0.36, -6.84) K = (-0.36, -6.84) L = (15, -6.84) L = (15, -6.84) L = (15, -6.84)
Vậy \(f\left(x\right)=0\) khi \(x=\dfrac{1}{2}\);
\(f\left(x\right)>0\) khi \(x\in\left(-2;-\dfrac{1}{2}\right)\cup\left(1;+\infty\right)\).
\(f\left(x\right)< 0\) khi \(x\in\left(-\infty;-2\right)\cup\left(-\dfrac{1}{2};1\right)\).

Đúng là giáo viên , level khác hẵng người thường

\(f\left(x\right)=\dfrac{3}{2x-1}-\dfrac{1}{x+2}=\dfrac{3\left(x+2\right)-\left(2x-1\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{x+7}{\left(2x-1\right)\left(x+2\right)}\).
\(x+7=0\Leftrightarrow x=-7\); \(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\); \(x+2=0\Leftrightarrow x=-2\).
TenAnh1 TenAnh1 B = (11.24, -6.26) B = (11.24, -6.26) B = (11.24, -6.26) C = (-0.38, -6.9) C = (-0.38, -6.9) C = (-0.38, -6.9) D = (14.98, -6.9) D = (14.98, -6.9) D = (14.98, -6.9) E = (-0.4, -6.68) E = (-0.4, -6.68) E = (-0.4, -6.68) F = (13.84, -6.58) F = (13.84, -6.58) F = (13.84, -6.58) G = (-0.44, -6.72) G = (-0.44, -6.72) G = (-0.44, -6.72) H = (14.92, -6.72) H = (14.92, -6.72) H = (14.92, -6.72)
Vậy \(f\left(x\right)=0\) khi \(x=\left\{-7\right\}\).
\(f\left(x\right)>0\) khi \(x\in\left(-7;-2\right)\cup\left(\dfrac{1}{2};+\infty\right)\).
\(f\left(x\right)< 0\) khi \(\left(-\infty;-7\right)\cup\left(-2;\dfrac{1}{2}\right)\).
\(f\left(x\right)\) không xác định tại \(x=\left\{\dfrac{1}{2};-2\right\}\)

Tết đến rồi ạ.

Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)

\(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)

\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)

\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{x}{1-x}=\frac{x}{\left(1-x\right)^2}\)

\(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)

\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)

\(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)

\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)

\(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

a: ĐKXĐ: x<>1; x<>-1

TH1: \(x^2-1>0\)

=>\(x^2>1\)

=>x>1 hoặc x<-1

\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)

\(=\frac{x+2}{\left(x^2-1\right)}+\frac{x^2}{x+1}\)

\(=\frac{x+2+x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^3-x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)

TH2: \(x^2-1<0\)

=>-1<x<1

\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)

\(=\frac{-\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x^2}{x+1}\)

\(=\frac{-\left(x+2\right)+x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^3-x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)

b: \(B=2x:\frac12x+\left(x+1\right)^2\)

\(=\left(2:\frac12\right)\cdot\left(\frac{x}{x}\right)+x^2+2x+1\)

\(=x^2+2x+1+4=x^2+2x+5\)

c: \(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)

\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)

\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{x}{1-x}=\frac{x}{\left(1-x\right)^2}\)

d: \(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)

\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)

e: \(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)

\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)

f: \(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

Mình nghĩ là dấu âm