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

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

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

          z-x=-12=> z=-12-x

  nên  y.z=\(-\frac{30}{x}.\left(-12-x\right)=42\)

             \(=\frac{360}{x}-\frac{30x}{x}=42\)

                \(=\frac{360-30x}{x}=42\)

              \(=>360-30x=42x\)

              \(=360-30x-42x=0\)

                   \(=360-72x=0\)

                    \(< =>72x=360\)

                          \(x=5\)=>  \(y=-6\);     \(z=-7\)

1.Ta có A= 7¹⁰ +7⁹  - 7⁸ 

A= 7⁸ .(7² +7 -1)

A=7⁸ .55

=> A chia hết cho 11( vì có thừa số 55 chia hết cho 11)

bạn biết làm câu 2 không

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

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)

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

\(=\dfrac{3}{x-y}\)

\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

\(=2\sqrt{3}-2+\sqrt{3}+1-3-\sqrt{3}\)

\(=2\sqrt{3}-4\)