c) Ta có: \(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\forall x\in Q\\\left|2,5-x\right|\ge0\forall x\in Q\end{matrix}\right.\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\forall x\in Q\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Vậy \(x=\left\{{}\begin{matrix}1,5\\2,5\end{matrix}\right.\).

e) \(\left(x-2\right)^2=1\)

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

Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\).

Mấy câu kia dễ rồi.

sửa lại ý c của N.Anh

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge1>0\)

mà theo đề thì \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\) k có gt \(x\) nào tm yêu cầu đề bài

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

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

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a/|x|-2,5=27,5

=>|x|=27,5+2,5=30

=>x=30 hoặc x=-30

b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)

=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)

c/(x-1)\(^5\)=-32

=>x-1=-2 vì (-2)\(^5\)=-32

=>x=-2+1=-1

d/\(\dfrac{4}{5}.x+0,5=4.5\)

=>\(\dfrac{4}{5}.x+0,5=20\)

=>\(\dfrac{4}{5}.x=20-0,5=19,5\)

=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)

Bài 2 :

a, x = \(\dfrac{-3}{-11}\) => x =\(\dfrac{3}{11}\)

=>| x | = \(\dfrac{3}{11}\)

=> x= \(\dfrac{3}{11}\) hoặc x = \(\dfrac{-3}{11}\)

Bài 3 :

a, | 4.(x-1)| =12

=> 4.(x-1)=12 hoặc 4.(x-1)=-12

\(\left[{}\begin{matrix}4.\left(x-1\right)=12\\4.\left(x-1\right)=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x-4=12\\4x-4=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x=16\\4x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy x = 4 hoặc x = -2

b,|2x+1|-5 =10

|2x+1|=15

=2x+1=15 hoặc 2x+=-15

+) 2x+1=15 = > 2x = 14 = > x =7

+)2x+1=-15 => 2x= -16 => x = -8

Vậy x=7 hoặc x = -8

c,|2,5-x|-1,3=0

|2,5-x|= 1,3

=>2,5 -x = 1,3 hoặc 2,5 - x = -1,3

+)2,5 - x = 1,3 => x = 1,2

+)2,5-x = -1,3 => x=3,8

Vậy x = 1,2 hoặc x = 3,8

d,-|1,4 - x | - 2 = 0

-|1,4-x|=2

=> -1,4+x = 2 hoặc -1,4+x = -2

+) -1,4+x= 2 => x = 3,4

+)-1,4+x= -2 => x = 0,6

Vậy x = 3,4 hoặc x = 0 ,6

e,| x - 2 | = x

=> x -2 = x hoặc x - 2 = -x

+) x- 2 = x => x-x = -2 => 0 = -2 ( vô lí )

+) x -2 = -x => x+x=2 => 2x =2 => x= 1

Vậy x = 1

f, 2.|2x-3| = \(\dfrac{1}{2}\)

=> |2x-3|= \(\dfrac{1}{4}\)

=>2x-3=\(\dfrac{1}{4}\) hoặc 2x-3=\(\dfrac{-1}{4}\)

+) 2x - 3 = \(\dfrac{1}{4}\)=> 2x= \(\dfrac{13}{4}\)=> x = \(\dfrac{13}{8}\)

+) 2x - 3 = \(\dfrac{-1}{4}\)=> 2x=\(\dfrac{11}{4}\)=> x = \(\dfrac{11}{8}\)

Vậy x=\(\dfrac{13}{8}\) hoặc x=\(\dfrac{11}{8}\)

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)