Ta có: \(4x+10⋮2x+3\)

\(\Rightarrow4x+6+4⋮2x+3\)

\(\Rightarrow4x+6⋮2x+3\)

\(\Rightarrow4⋮2x+3\)

\(\Rightarrow2x+3\inƯ_{\left(4\right)}=-4;-2;-1;1;2;4\)

* Nếu: \(2x+3=-4\Rightarrow2x=-7\Rightarrow x=\frac{-7}{2}\)

* Nếu: \(2x+3=-2\Rightarrow2x=-5\Rightarrow x=\frac{-5}{2}\)

* Nếu: \(2x+3=-1\Rightarrow2x=-4\Rightarrow x=-2\)

* Nếu: \(2x+3=1\Rightarrow2x=-2\Rightarrow x=-1\)

* Nếu: \(2x+3=2\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)

* Nếu: \(2x+3=4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Vậy: \(x=-\frac{7}{2};-\frac{5}{2};-2;-1;-\frac{1}{2};\frac{1}{2}\)

Ta có: \(2x+6⋮x+2\)

\(\Rightarrow2x+4+2⋮x+2\)

\(\Rightarrow2x+4⋮x+2\)

\(\Rightarrow2⋮x+2\)

\(\Rightarrow x+2\inƯ_{\left(2\right)}=-2;-1;1;2\)

\(\Rightarrow\left[\begin{matrix}x+2=-2\\x+2=-1\\x+2=1\\x+2=2\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-4\\x=-3\\x=-1\\x=0\end{matrix}\right.\)

Vậy: \(x=-4;-3;-1;0\)

câu b: 5x=10 hay 5x+10

duong thang MN, MP,MQ,NP.NQ,PQ . 6 duong thang

Ta có: \(x^2\left(x+2\right)+4\left(x+2\right)=0\)

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

\(\Rightarrow x+2=0\) vì \(x^2+4>0\)

\(\Rightarrow x=-2\)

Vậy: \(x=-2\)

mk vẽ ko ẹp lắm

thông cảm

3) 3a - b - 2c = 2 với b = 6 ; c = -1

Thay số: \(3a-6-\left(2.-1\right)=2\)

\(\Rightarrow3a-6-2=2\)

\(\Rightarrow3a-8=2\)

\(\Rightarrow3a=10\)

\(\Rightarrow a=\frac{10}{3}\)

Vậy: \(a=\frac{10}{3}\)

4) 12 - a + b + 5c = -1 với b = -7 ; c = 5

Thay số: \(12-a+\left(-7\right)+5.5=-1\)

\(\Rightarrow12-a-7+25=-1\)

\(\Rightarrow12-a+18=-1\)

\(\Rightarrow12+18-a=-1\)

\(\Rightarrow20-a=-1\)

\(\Rightarrow a=21\)

Vậy: \(a=21\)

5) 1 - 2b + c - 3a =-9 với b = -3 ; c = -7

Thay số: \(1-\left(2.-3\right)+\left(-7\right)-3a=-9\)

\(\Rightarrow1--6-7-3a=-9\)

\(\Rightarrow1+6-7-3a=-9\)

\(\Rightarrow7-7-3a=-9\)

\(\Rightarrow-3a=-9\)

\(\Rightarrow a=3\)

Vậy: \(a=3\)

còn 2 câu trên hình như thiếu đề á!!!