\(B=\frac{35^3+5.35^2-5^3.7}{10.70^2+10^2.70-10^3}=\frac{5^3.7^3+5^3.7^2-5^3.7}{10^3.7^2+10^3.7-10^3}=\frac{5^3.7.\left(7^2+7-1\right)}{10^3.\left(7^2+7-1\right)}.\)

=> \(B=\frac{5^3.7}{10^3}=\frac{5^3.7}{2^3.5^3}=\frac{7}{2^3}=\frac{7}{8}\)

a) ta có : \(z-x=-12\Leftrightarrow z=x-12\)

\(\Rightarrow yz=42\Leftrightarrow y\left(x-12\right)=42\Leftrightarrow xy-12y=42\)

\(\Leftrightarrow-30-12y=42\Leftrightarrow12y=-30-42=-72\Leftrightarrow y=\dfrac{-72}{12}=-6\)

ta có : \(y=-6\Rightarrow xy=-30\Leftrightarrow x.-6=-30\Leftrightarrow x=\dfrac{-30}{-6}=5\)

ta có : \(x=5\Rightarrow z=5-12=-7\)

vậy \(x=5;y=-6;z=-7\)

b) ta có :\(A=7^{10}+7^9-7^8=7^8.\left(7^2+7-1\right)=7^8.55=7^8.5.11⋮11\)

\(\Leftrightarrow7^8.5.11\) chia hết cho \(11\) \(\Leftrightarrow\) A chia hết cho 11

vậy A chia hết cho 11 (đpcm)

a)xy=30 ;yz=42=>\(y=\dfrac{30}{x}\);\(y=\dfrac{42}{z}\)

Do đó \(\dfrac{30}{x}=\dfrac{42}{z}\)

Áp dụng t/c của dãy tỉ số bằng nhau,tac có:

\(\dfrac{30}{x}=\dfrac{42}{z}\)=\(\dfrac{42-30}{z-x}\)=\(\dfrac{12}{-12}=-1\)

=>x=-30;z=-42

Do đó y=\(\dfrac{30}{x}=\dfrac{30}{-30}=-1\)

)xy=-30 ;yz=42=>y=30x;y=42z

Do đó -30/x=42/z

Áp dụng t/c của dãy tỉ số bằng nhau,tac có:

-30/x=42/z
=42+30/z−x=72/−12=−6

=>x=-5;z=--7

Do đó y=30/x=30/-5=−6

Bài 2: 

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

Ta có: xy=12

\(\Leftrightarrow12k^2=12\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)

Giải:

Ta có:

x.y=-30 => \(\frac{x}{-30}=\frac{1}{y}\)(1)

y.z=42 => \(\frac{z}{42}=\frac{1}{y}\)(2)

Từ (1) và (2) => \(\frac{x}{-30}=\frac{z}{42}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta được:

\(\frac{x}{-30}=\frac{z}{42}=\frac{z-x}{42-\left(-30\right)}=\frac{12}{72}=\frac{1}{6}\)

\(\frac{x}{-30}=\frac{1}{6}\)=> x=-5

\(\frac{z}{42}=\frac{1}{6}\)=>z=7

Thay x=-5 vào x.y=-30 ta được:

-5.y=-30=>y=-6

Vậy x=-5;y=-6;z=7

Đúng thì k mình nhé!!!

\(B=\dfrac{35^3+5\cdot35^2-5^3\cdot7}{10\cdot70^2+10^2\cdot70-10^3}=\dfrac{\left(5\cdot7\right)^3+5\cdot\left(5\cdot7\right)^2-5^3\cdot7}{2\cdot5\cdot\left(2\cdot5\cdot7\right)^2+\left(2\cdot5\right)^2\cdot2\cdot5\cdot7-\left(2\cdot5\right)^3}=\dfrac{5^3\cdot7^3+5\cdot5^2\cdot7^2-5^3\cdot7}{2\cdot5\cdot2^2\cdot5^2\cdot7^2+2^2\cdot5^2\cdot2\cdot5\cdot7-2^3\cdot5^3}=\dfrac{5^3\cdot7^3+5^3\cdot7^2-5^3\cdot7}{2^3\cdot5^3\cdot7^2+2^3\cdot5^3\cdot7-2^3\cdot5^3}=\dfrac{5^3\left(7^3+7^2-7\right)}{2^3\cdot5^3\left(7^2+7-1\right)}=\dfrac{343+49-7}{8\cdot\left(49+7-1\right)}=\dfrac{385}{8\cdot55}=\dfrac{385}{440}=\dfrac{7}{8}\)

Vậy \(B=\dfrac{7}{8}\)

Ta có:\(xy-yz\)=-30-42

=>\(y\left(x-z\right)\)=-72

=>-12\(y\)=-72

=>\(y\)=6

vì \(xy=-30\)

\(\Leftrightarrow\) \(x.6=-30\)

\(\Leftrightarrow\)\(x=-5\)

Vì \(z-x=-12\)

\(\Leftrightarrow\)\(z-\left(-5\right)=-12\)

\(\Leftrightarrow\)\(z+5=-12\)

\(\Leftrightarrow\)\(z=-17\)

VẬY \(\left(x;y;z\right)=\left(6;-5;-17\right)\)

suy ra:  (x.y)/(y.z)=-30/42 suy ra x/z=-5/7 suy ra x=-5;z=7 suy ra y=-30:-5=6          

       Vậy x=-5 ; z=7 ;y=6

\(x-y=-30\Rightarrow\dfrac{x}{-30}=\dfrac{1}{y}\\ y.z=-42\\ \Rightarrow\dfrac{z}{-42}=\dfrac{1}{y}\\ \Rightarrow\dfrac{x}{-30}=\dfrac{z}{-42}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{-30}=\dfrac{z}{-42}=\dfrac{z-x}{-42-\left(-30\right)}=\dfrac{-12}{-12}=1\)

\(\dfrac{x}{-30}=1\Rightarrow x=-30\\ \dfrac{z}{-42}=1\Rightarrow z=-42\)

\(x.y=-30\Rightarrow-30.y=-30\Rightarrow y=1\)

\(xy=-30;yz=42\)

\(xy-yz=-72\) => \(y\left(x-z\right)=-72\) => \(y.\left(-12\right)=-72\) => \(y=-\frac{72}{-12}=6\) 

\(xy=-30\) => \(x=-5\)

\(yz=42\) => \(z=7\)