\(\Rightarrow2x-x=-2-2\Rightarrow x=-4\)

Bài 4:

a) Áp dụng t/c dtsbn:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{14}{7}=2\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=2.3=6\end{matrix}\right.\)

Vậy....

b) Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{2x}{16}=\dfrac{3y}{36}=\dfrac{2x+3y}{16+36}=\dfrac{13}{52}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}.8=2\\y=\dfrac{1}{4}.12=3\end{matrix}\right.\)(nhận)

Vậy...

Nửa chu vi là: \(120:2=60\left(cm\right)\)

Gọi độ dài 2 cạnh đó lần lượt là a,b(cm)(a>b>0)

Áp dụng t/c dtsbn:

\(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{a+b}{4+3}=\dfrac{60}{7}\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{60}{7}.4=\dfrac{240}{7}\\b=\dfrac{60}{7}.3=\dfrac{180}{7}\end{matrix}\right.\)(nhận)

Vậy....

\(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

Điện trở tương đương của đoạn mạch là:

\(R_{tđ}=R_1+R_2=3+9=12\left(\Omega\right)\Rightarrow B\)

C

\(3x-2< 10\Leftrightarrow3x< 12\Leftrightarrow x< 4\)

Mà \(x\in N\Rightarrow x\in\left\{0;1;2;3\right\}\Rightarrow D\)

4) \(\Rightarrow\left(3x+7\right)\left(x+5\right)-\left(x+5\right)\left(3-2x\right)=0\)

\(\Rightarrow\left(x+5\right)\left(5x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{4}{5}\end{matrix}\right.\)

5) \(\Rightarrow2x\left(x-3\right)\left(x+3\right)-5\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+3\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)