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) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

a: =>2sin(x+pi/3)=-1

=>sin(x+pi/3)=-1/2

=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi

=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi

b: =>2sin(x-30 độ)=-1

=>sin(x-30 độ)=-1/2

=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ

=>x=k*360 độ hoặc x=240 độ+k*360 độ

c: =>2sin(x-pi/6)=-căn 3

=>sin(x-pi/6)=-căn 3/2

=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi

=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi

d: =>2sin(x+10 độ)=-căn 3

=>sin(x+10 độ)=-căn 3/2

=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ

=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ

e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)

=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)

=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ

=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ

f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)

=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi

=>x=pi/12+k2pi hoặc x=19/12pi+k2pi

g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

h) \(1+sin\left(x-30^o\right)=0\)

\(\Leftrightarrow sin\left(x-30^o\right)=-1\)

\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow x=-60^0+k360^o\)

a)

\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

@Hùng Nguyễn

| 4x - 0,2 | = 0,2

=> 4x - 0,2 = \(\dfrac{-1}{5}\)

4x = \(\dfrac{-1}{5}\) + 0,2 = 0

x = \(\dfrac{0}{4}\)

=> Ko có giá trị x

=> 4x - 0,2 = \(\dfrac{1}{5}\)

4x = \(\dfrac{1}{5}\)+ 0,2 = \(\dfrac{2}{5}\)

x = \(\dfrac{2}{5}\): 4 = \(\dfrac{2}{5}\). 4 = \(\dfrac{8}{5}\)

Vậy x = \(\dfrac{8}{5}\)

a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)

\(=1-x\sqrt{x}-x\sqrt{x}\)

\(=1-2x\sqrt{x}\)

b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\left(\dfrac{\left(1-\sqrt{a}\right)\cdot\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\)

\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

mình làm lại câu b) nha

b) |x-3|=-4

th1: x-3=-4

x=3+(-4)

x=-1

th2: x-3=4

x=3+4

x=7

b) \(\left|x-3\right|=-4\)

t/h1:\(x-3=-4\)

\(x=3-\left(-4\right)\)

\(x=7\)

t/h2:\(x-3=4\)

\(x=3-4\)

\(x=-1\)

c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).

Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).

b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).

Đẳng thức xảy ra khi x = \(\sqrt{6}\).

Câu a muốn có min thì đề bài phải là \(x\ge4\) (có dấu "=")

Còn \(x>4\) thì chắc là đề sai