\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

\(a.\)\(\frac{13x-16}{15}+\frac{x-32}{35}< \frac{x-6}{21}\)\(MC:105\)

\(\Leftrightarrow\frac{7\left(13x-16\right)}{105}+\frac{3\left(x-2\right)}{105}< \frac{5\left(x-6\right)}{105}\)

\(\text{Khử mẫu ta dc pt tương đương vs pt:}\)

\(\Leftrightarrow7\left(13x-16\right)+3\left(x-2\right)< 5\left(x-6\right)\)

\(\Leftrightarrow91x-112+3x-6< 5x-30\)

\(\Leftrightarrow94x-118< 5x-30\)

\(\Leftrightarrow94x-5x< 118-30\)

\(\Leftrightarrow89x< 88\)

\(\Leftrightarrow x< \frac{88}{89}\)

.\(b.\)\(\frac{5x+12}{14}+\frac{11x+28}{3}>\frac{4x+9}{17}\)\(MC:714\)

\(\text{Khi khử mẫu pt ta dc pt tương đương}:\):

\(\Leftrightarrow51\left(5x+12\right)+238\left(11x+28\right)>42\left(4x+9\right)\)

\(\Leftrightarrow255x+612+2618x+6664>168x+378\)

\(\Leftrightarrow2873x+7276>168x+378\)

\(\Leftrightarrow2873x-168x>-7276+378\)

\(\Leftrightarrow2705x>-6898\)

\(\Leftrightarrow x>-\frac{6898}{2705}\)

- \(x^2\) + 5\(x\) - 4 = 0

-\(x^2\) + \(x\) + 4\(x\) - 4 = 0

(- \(x^2\) + \(x\)) + (4\(x\) - 4) = 0

-\(x\)(\(x-1\)) + 4\(\times\)( \(x\) -1) = 0

(\(x-1\))( -\(x\) +4) = 0

\(\left[{}\begin{matrix}x-1=0\\-x+4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

\(x\) \(\in\) { 1; 4}

`-x^2+5x-4 =0`

`\Rightarrow x^2-5x+4=0`

`\Rightarrow x^2-4x-x+4=0`

`\Rightarrow (x^2-4x)-(x-4)=0`

`\Rightarrow x(x-4)-(x-4)=0`

`\Rightarrow (x-4)(x-1)=0`

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=0+4\\x=0+1\end{matrix}\right.\)

``\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

Vậy, nghiệm của đa thức là `x={4; 1}.`

25 . (52 - x) = 5

=> 52 - x = 5 : 25

=> 52 - x = 0,2

=> x = 52 - 0,2

=> x = 51,8

\(25.\left(52-x\right)=5\)

\(52-x=5:25\)

\(52-x=0,2\)

\(x=52-0,2\)

\(x=51,8\)

Ta có \(x\inƯ\left(30\right)\)\(\left(ĐKXĐ:x\le8\right)\)

\(< =>x\in\left\{1;2;3;5;6;10;15;30\right\}\)

Do \(x\le8\)suy ra ta có bộ số x thỏa mãn sau :

\(x\in\left\{1;2;3;5;6\right\}\)

Trả lời :

Theo bài ta có :

\(30⋮x\Rightarrow x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)

Mà \(x< 8\Rightarrow x\in\left\{1;2;3;5;6\right\}\)

Cảm ơn các bạn nhé

15 - 2 | 4 + 5x | = 59 <=> 2 | 4 + 5x | = -44 <=> | 4 + 5x | = -22

Vì \(\left|4+5x\right|\ge0\) nên không có x thỏa mãn | 4 + 5x | = -22

-------------------------------

TH1: \(x\le\frac{1}{5}\)=>3-x+1-5x=7=>4-6x=7=>-3=6x=>x=-1/2(nhận)

TH2:\(\frac{1}{5}< x\le3\)=>3-x+5x-1=7=>2+4x=7=>4x=5=>x=5/4(nhận)

TH3:x>3=>x-3+5x-1=7=>6x-4=7=>6x=11=>x=11/6(loại)

Vậy x=-1/2 hoặc x=5/4