a) \(\frac{3}{4}-\left|2x+1\right|=\frac{7}{8}\)

\(\left|2x+1\right|=\frac{3}{4}-\frac{7}{8}\)

\(\left|2x+1\right|=-\frac{1}{8}\)

\(\Rightarrow x\in\varnothing\)

b) \(2.\left|2x-3\right|=\frac{1}{2}\)

\(\left|2x-3\right|=\frac{1}{4}\)

TH1: 2x - 3 = 1/4

...

TH2: 2x -3  = -1/4

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rùi bn tự lm típ nhé! câu c dựa vào phần a;b là lm đk

d)\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=1,6\)

...

a) \(\left|x-1,7\right|+\dfrac{1}{2}=23\)

\(\left|x-1,7\right|=22,5\)

\(\Rightarrow x-1,7=\pm22,5\)

\(\Rightarrow x=\left\{24,2;-20,8\right\}\)

b) \(\left|x+\dfrac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)

\(\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\dfrac{4}{15}\right|=1,6\)

\(\Rightarrow x+\dfrac{4}{15}=\pm1,6\)

\(\Rightarrow x=\left\{\dfrac{4}{3};-\dfrac{28}{15}\right\}\)

a)\(\left|x-1,7\right|+\dfrac{1}{2}=23\)

<=>\(\left|x-1,7\right|=22,5\)

<=>\(-\left(x-1,7\right)=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x+1,7=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x=-22,5-1,7\) hoặc \(x=22,5+1,7\)

<=>\(x=24,2\)

vậy \(x=24,2\)

a. \(\dfrac{3}{4}-\left|2x+1\right|=\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{-1}{8}\)

=> \(\left\{{}\begin{matrix}2x+1=\dfrac{-1}{8}\\2x+1=\dfrac{1}{8}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{-9}{16}\\x=\dfrac{-7}{16}\end{matrix}\right.\)

#Yiin

b. \(2.\left|2x-3\right|=\dfrac{1}{2}\)

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

=> \(\left\{{}\begin{matrix}2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)

\(P\left(x\right)=2x^4+3x^2-x^3-3x^4-x^2-2x+1\)

\(=-x^4-x^3+2x^2-2x+1\)

C

`D(x)=2x^4+7x^2=0`

`-> x(2x^3+7x)=0`

`->`\(\left[{}\begin{matrix}x=0\\2x^3+7x=0\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\x\left(2x^2+7\right)=0\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\x=0\\2x^2+7=0\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\2x^2=-7\text{ }\left(\text{k t/m}\right)\end{matrix}\right.\)

Vậy, nghiệm của đa thức là `x=0`

`E(x)=8x^4+x=0`

`-> x(8x^3+1)=0`

`->`\(\left[{}\begin{matrix}x=0\\8x^3+1=0\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\8x^3=-1\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\x^3=-\dfrac{1}{8}\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy, nghiệm của đa thức là `x={0 ; -1/2}`

`F(x)=x(-2x+3)+2x^2-5=0`

`-> -2x^2+3x+2x^2-5=0`

`-> 3x-5=0`

`-> 3x=5`

`-> x=5/3`

Vậy, nghiệm của đa thức là `x=5/3`.

cảm ơn cậu raastttttttttttttttttttttt nhiều

Lời giải:

a.

$|2x-5|=12-3x$

Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$

$\Leftrightarrow x=3,4$ (thỏa mãn)

Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$

$\Leftrightarrow x=7$ (loại)

Vậy......

b.

$4x=|x+1|+|x+2|+|x+3|\geq 0$

$\Rightarrow x\geq 0$

Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$

Vậy: $3x+6=4x$

$\Leftrightarrow x=6$ (thỏa mãn)

c.

$|x^2+|x+2||=x^2+3$

$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$

$\Leftrightarrow x+2=3$ hoặc $x+2=-3$

$\Leftrightarrow x=1$ hoặc $x=-5$

d.

$|x^2-3|=6$

$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$

$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)

$\Leftrightarrow x=\pm 3$

a/ 3(1 - x) - 5(2x - 2) = 0 

    => 3 - 3x - 10x + 10 = 0

    => -13x = -13

    => x = 1

     Vậy x = 1

b/ |3x - 2| - 4 = 0 => |3x - 2| = 4  

    Suy ra 2 trường hợp:

•      3x - 2 = 4 => 3x = 6 => x = 2
•      3x - 2  = -4 => 3x = -2 => x = -2/3

    Vậy x = 2 , x = -2/3

c/ 2x - x³ = 0 => x.(2 - x²) = 0 

     => x = 0 

    hoặc 2 - x² = 0 => x² = 2 => x = \(\sqrt{2}\)  hoặc x = \(-\sqrt{2}\)

         Vậy \(x=\left\{0;\sqrt{2};-\sqrt{2}\right\}\)

d/ x(1 - 2x) + (2x² - x + 4) = 0 

     => x - 2x² + 2x² - x + 4 = 0

     => 4 = 0 (vô lí)

     Vậy vô nghiệm