a)

\(A=\left(2x\right)^3-1-7x^3-7=8x^3-1-7x^3-7=x^3-8\)

b)

Thay \(x=-\dfrac{1}{2}\) vào biểu thức A:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{65}{8}\)

13: \(\frac{2x+3}{M}=\frac{4x^2-9}{10x^2-30x}\)

=>\(\frac{2x+3}{M}=\frac{\left(2x-3\right)\left(2x+3\right)}{10x\left(x-3\right)}\)

=>\(\frac{1}{M}=\frac{2x-3}{10x\left(x-3\right)}\)

=>\(M=\frac{10x\left(x-3\right)}{2x-3}=\frac{10x^2-30x}{2x-3}\)

giúp bài j

:v

Em chụp lại c8

đây ạ

105,48 x 1,6 x (1,25 x 5 - 1,25 : 0,2)   
= 168,768 x 0 = 0

\(105,48\times1,6\times\left(1,25\times5-1,25\div0,2\right)\)

\(=168,768\times\left(6,25-6,25\right)\)

\(=168,768\times0\\ =0\)

Giúp mình bài 1 ,2,3,4, 5 , 6, 7 ,8 ạ

\(\left\{{}\begin{matrix}3x+5y=1\\x-8y=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\3x-24y=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}29y=-29\\x-8y=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-8\left(-1\right)=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;-1\right)\)

\(\left\{{}\begin{matrix}3x+5y=1\\3x+5y=1\end{matrix}\right.\)

Có sai dấu ko bạn 

Nếu lấy pt thứ nhất trừ thứ 2 ra 0 là vô lý r

Ta có: \(\frac{a+b}{b+c}=\frac79\)

=>\(\frac{a+b}{7}=\frac{b+c}{9}\)

\(\frac{c+a}{a+b}=\frac87\)

=>\(\frac{a+c}{8}=\frac{a+b}{7}\)

=>\(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}\)

Đặt \(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}=k\)

=>a+b=7k; a+c=8k; b+c=9k

=>a+c-a-b=8k-7k=k; c+b=9k; a+b=7k

=>c-b=k và c+b=9k và a+b=7k

=>c=(k+9k)/2=5k; b=9k-5k=4k; a=7k-4k=3k

\(P=\frac{3ab+2bc+ac}{a^2+2b^2+3c^2}\)

\(=\frac{3\cdot3k\cdot4k+2\cdot4k\cdot5k+3k\cdot5k}{\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2}\)

\(=\frac{36k^2+40k^2+15k^2}{9k^2+32k^2+75k^2}=\frac{36+40+15}{9+32+75}=\frac{91}{116}\)

Ta có: \(\frac{a+b}{b+c}=\frac79\)

=>\(\frac{a+b}{7}=\frac{b+c}{9}\)

\(\frac{c+a}{a+b}=\frac87\)

=>\(\frac{a+c}{8}=\frac{a+b}{7}\)

=>\(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}\)

Đặt \(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}=k\)

=>a+b=7k; a+c=8k; b+c=9k

=>a+c-a-b=8k-7k=k; c+b=9k; a+b=7k

=>c-b=k và c+b=9k và a+b=7k

=>c=(k+9k)/2=5k; b=9k-5k=4k; a=7k-4k=3k

\(P=\frac{3ab+2bc+ac}{a^2+2b^2+3c^2}\)

\(=\frac{3\cdot3k\cdot4k+2\cdot4k\cdot5k+3k\cdot5k}{\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2}\)

\(=\frac{36k^2+40k^2+15k^2}{9k^2+32k^2+75k^2}=\frac{36+40+15}{9+32+75}=\frac{91}{116}\)