a: \(\left|x+\frac{19}{55}\right|\ge0\forall x\)

\(\left|y+\frac{1890}{1975}\right|\ge0\forall y\)

\(\left|z-2004\right|\ge0\forall z\)

Do đó: \(\left|x+\frac{19}{55}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac{19}{55}=0\\ y+\frac{1890}{1975}=0\\ z-2004=0\end{cases}\Rightarrow\begin{cases}x=-\frac{19}{55}\\ y=-\frac{1890}{1975}=-\frac{378}{395}\\ z=2004\end{cases}\)

b: Sửa đề: \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\le0\)

Ta có: \(\left|x+\frac92\right|\ge0\forall x\)

\(\left|y+\frac43\right|>=0\forall y\)

\(\left|z+\frac72\right|\ge0\forall z\)

Do đó: \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\ge0\forall x,y,z\)

mà \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\le0\)

nên \(\begin{cases}x+\frac92=0\\ y+\frac43=0\\ z+\frac72=0\end{cases}\Rightarrow\begin{cases}x=-\frac92\\ y=-\frac43\\ z=-\frac72\end{cases}\)

c: \(\left|x+\frac34\right|\ge0\forall x\)

\(\left|y-\frac15\right|\ge0\forall y\)

\(\left|x+y+z\right|\ge0\forall x,y,z\)

Do đó: \(\left|x+\frac34\right|+\left|y-\frac15\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac34=0\\ y-\frac15=0\\ x+y+z=0\end{cases}\Rightarrow\begin{cases}x=-\frac34\\ y=\frac15\\ z=-x-y=\frac34-\frac15=\frac{11}{20}\end{cases}\)

d: \(\left|x+\frac34\right|\ge0\forall x\)

\(\left|y-\frac25\right|\ge0\forall y\)

\(\left|z+\frac12\right|\ge0\forall z\)

Do đó: \(\left|x+\frac34\right|+\left|y-\frac25\right|+\left|z+\frac12\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac34=0\\ y-\frac25=0\\ z+\frac12=0\end{cases}\Rightarrow\begin{cases}x=-\frac34\\ y=\frac25\\ z=-\frac12\end{cases}\)

X:(\(\frac{2}{9}-\frac{1}{5}\))=\(\frac{8}{16}\)

x:\(\frac{1}{45}\) =\(\frac{8}{16}\)

x: =\(\frac{8}{16}.\frac{1}{45}\)

x: =\(\frac{1}{90}\)

a) \(\sqrt{x^2-4x+4}=\sqrt{\left(x-2\right)^2}=3\Leftrightarrow x-2=3\Leftrightarrow x=5\)

b) \(\sqrt{x^2-12}=2\) \(\Leftrightarrow x^2-12=4\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

c) \(\sqrt{x+3}=x+3\Leftrightarrow x+3-\sqrt{x+3}=0\)

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

mấy câu còn lại bn làm tương tự

Mysterious Person Akai Haruma

a: \(\left(x^2-3\right)\left(2x^2-\dfrac{9}{8}\right)\left(\sqrt{\left|x\right|}-\sqrt{\dfrac{5}{2}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=0\\2x^2-\dfrac{9}{8}=0\\\sqrt{\left|x\right|}-\sqrt{\dfrac{5}{2}}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=3\\x^2=\dfrac{9}{16}\\\left|x\right|=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow x\in\left\{-\sqrt{3};\sqrt{3};\dfrac{3}{4};-\dfrac{3}{4};\dfrac{-5}{2};\dfrac{5}{2}\right\}\)

b: \(x-5\sqrt{x}=0\)(ĐKXĐ: x>=0)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)

=>x=0 hoặc x=25

\(b.\) \(\left(x-1\right).\left(x-2\right)>0\)

\(\Leftrightarrow x-1\) và \(x-2\) cùng dấu

\(\Leftrightarrow\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\)     Hoặc: \(\Leftrightarrow\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)

T/hợp 1:   \(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\)

T/hợp 2: \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}}\)

Vậy: ..................................

\(e.\)\(\frac{5}{x}< 1\)

\(\Leftrightarrow x>5\)

Vậy: .............................

a) x² - 9 + (x + 3) = 0

=> (x - 3).(x + 3) + (x + 3) = 0

=> (x + 3).(x - 3 + 1) = 0

=> (x + 3).(x - 2) = 0

=> \(\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

b) x² - 5x + 6 = 0

=> x² - 2x - 3x + 6 = 0

=> x.(x - 2) - 3.(x - 2) = 0

=> (x - 2).(x - 3) = 0

=> \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(x^2-9+\left(x+3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+3\right)+\left(x+3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}}\)

\(x^2-5x+6=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

a: \(\left(x-2\right)^2\cdot\left(x+1\right)\left(x-4\right)< 0\)

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

=>-1<x<4

b: \(\dfrac{x^2\left(x-3\right)}{x-9}< 0\)

\(\Leftrightarrow\dfrac{x-3}{x-9}< 0\)

=>3<x<9

a: 1-2x<7

=>-2x<6

hay x>-3

b: (x-1)(x-2)>0

=>x-2>0 hoặc x-1<0

=>x>2 hoặc x<1

c: \(\left(x-2\right)^2\cdot\left(x+1\right)\left(x-4\right)< 0\)

=>(x+1)(x-4)<0

=>-1<x<4