a)

Ta có: \(9x=5y=15z\Rightarrow\dfrac{9x}{45}=\dfrac{5y}{45}=\dfrac{15z}{45}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{3}\Rightarrow\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}_{\left(1\right)}\)

và \(-x+y-z=11_{\left(2\right)}.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ só bằng nhau có:

\(\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}=\dfrac{-x+y-z}{-5+9-3}=\dfrac{11}{1}=11.\)

Từ đó: \(\left\{{}\begin{matrix}\dfrac{-x}{-5}=11\Rightarrow-x=-55\Rightarrow x=55.\\\dfrac{y}{9}=11\Rightarrow y=99.\\\dfrac{z}{3}=11\Rightarrow z=33.\end{matrix}\right.\)

Vậy.....

b); c); d); e) làm tương tự.

Bài 1:

Giải:

Ta có: \(\left\{{}\begin{matrix}3x=4y\\5y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Mà \(xyz=30\)

\(\Rightarrow240k^3=30\)

\(\Rightarrow k^3=\dfrac{1}{8}\)

\(\Rightarrow k=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=2,5\end{matrix}\right.\)

Vậy...

Bài 2: sai đề

Bài 3:

Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Ta có: \(x+2y+3z=38\)

\(\Rightarrow2k+1+8k-6+18k+15=38\)

\(\Rightarrow28k=28\)

\(\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\\z=11\end{matrix}\right.\)

Vậy...

1) Ta có :

\(3x=4y\Rightarrow\dfrac{3x}{12}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\) <=> \(\dfrac{x}{8}=\dfrac{y}{6}\)

\(5y=6z\Rightarrow\dfrac{5y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\)

=> \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Thay vào đẳng thức xyz = 30

=> 8k.6k.5k = 30

<=> 240k³ = 30

<=> k³ = 8

<=> k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)

b) Câu này cũng tương tự câu 1 nha ! Đặt k luôn , còn không bình phương lên rồi dùng tính chất dãy tỉ số bằng nhau .

c) Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)

=> \(\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Thay vào đẳng thức , ta có :

x + 2y + 3z = 2k + 1 + 2(4k - 3) + 3(6k + 5) = 38

=> 28k = 38

=> k = \(\dfrac{19}{14}\)

Vậy .....

Câu hỏi của Sương Đặng - Toán lớp 7 | Học trực tuyến

\(A=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-b\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left(1-b\right)}\)

\(B=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)

\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)

\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)

đề sai

Đề sai rồi bạn

\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)