Question

[Phạm Đức Anh]

bài 1 tính

\(A=\frac{a+b}{b+c}\) biết \(\frac{b}{a}=2;\frac{c}{b}=3\)

bài 2 tìm x

a) \(\frac{72-x}{7}=\frac{x-40}{9}\)

b) \(\frac{x+4}{20}=\frac{5}{x+4}\)

bài 3 tìm x,y

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

bài 8 tìm x,y,z

a) x:y:z=3:4:5 và 2x²+2y²-3z²=-100

b)\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)

c) \(\left|x-3\right|+\left|y+5\right|+\left|x+y+z\right|=0\)

d) \(\left|2x-5\right|+\left|2y-z\right|+\left|4z-2\right|=0\)

Answer 1

[Vũ Minh Tuấn]

Bài 1:

\(A=\frac{a+b}{b+c}.\)

Ta có:

\(\frac{b}{a}=2\Rightarrow\frac{b}{2}=\frac{a}{1}\) (1)

\(\frac{c}{b}=3\Rightarrow\frac{c}{3}=\frac{b}{1}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{b}{2}=\frac{c}{6}.\)

\(\Rightarrow\frac{a}{1}=\frac{b}{2}=\frac{c}{6}=\frac{a+b}{3}=\frac{b+c}{8}.\)

\(\Rightarrow A=\frac{a+b}{b+c}=\frac{3}{8}\)

Vậy \(A=\frac{a+b}{b+c}=\frac{3}{8}.\)

Bài 2:

a) \(\frac{72-x}{7}=\frac{x-40}{9}\)

\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow648+280=7x+9x\)

\(\Rightarrow928=16x\)

\(\Rightarrow x=928:16\)

\(\Rightarrow x=58\)

Vậy \(x=58.\)

b) \(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow x+4=\pm10.\)

\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

Vậy \(x\in\left\{6;-14\right\}.\)

Chúc bạn học tốt!

Answer 2

[Diệu Huyền]

Bài 2:

a, \(\frac{72-x}{7}=\frac{x-40}{9}\)

\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)

\(\Rightarrow9.72-9.x=7.x-7.40\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow-9x-7x=-280-648\)

\(\Rightarrow-16x=-648\)

\(\Rightarrow x=58\)

Vậy \(x=58\)